Mathematics - 2020-09-18 (page 1 of 2)

Anindya Prithvi

04:30

Cool! Thankss, btw any ideas on how to generalize the above for
$$\dfrac{{xn}\choose{(x-2)n}}{{(x-1)n}\choose{(x-3)n}}$$ ....For x=3, I got 27/16...and x=4 gives 64/27 ....kind of seeing a pattern

Sayan Chattopadhyay

07:23

Representation theory is annoying. Why does the regular representation encode any information about the irreducible representations of $G$? To me both seem completely separate entities. (I do get proof-wise how you would show it, but I don't have any intuition whatsoever)

Tobias Kildetoft

07:38

@SayanChattopadhyay Well, it encodes the information because it contains all of the irreducibles

Sayan Chattopadhyay

@TobiasKildetoft I get that, but why should I expect it to?

Tobias Kildetoft

I am not sure there is a completely intuitive reason

In some sense it is because the fact that the regular representation contains the identity element of the group, you get to map to any representation you want to via a non-trivial map

Another is that the regular representation is induced from the trivial subgroup

Leaky Nun

@SayanChattopadhyay because any module is a quotient of $\Bbb CG^n$

so if you think of modules as numbers and irreducible modules as prime numbers

Sayan Chattopadhyay

Ah okay, I see

Leaky Nun

then every number is a divisor of a big enough power of $\Bbb CG$

so $\Bbb CG$ has to contain all the primes

@TobiasKildetoft there's something special about $\Bbb CG$ that makes this work

for example it doesn't work with $\Bbb Z$ because $\Bbb Z/2\Bbb Z$ is a simple module that isn't a direct summand of $\Bbb Z$

Tobias Kildetoft

07:49

@LeakyNun Yes, we need to be careful to not mix up the left- and the right induced modules in the general case

and hence not mix up whether we get submodules or quotients

Leaky Nun

well $\Bbb Z$ is commutative

Tobias Kildetoft

That does not make then coincide

Sayan Chattopadhyay

Oh so if I am doing it with the algebraic closure of some other field this won't work?

Leaky Nun

what's the general statement, if any?

@SayanChattopadhyay no the same proof works with any algebraically closed field with characteristic 0

Tobias Kildetoft

Not sure what the precise general statement would be for infinite groups

Leaky Nun

07:51

(you need characteristic 0 to divide by the order of the group)

Tobias Kildetoft

Usually you add some topological stuff to make things behave nicely in those cases

Leaky Nun

or more generally, with characteristic not dividing the order of the group

@TobiasKildetoft I'm talking about irreducible modules over a general ring

Tobias Kildetoft

Ahh, so the question is what makes group algebras special in this case?

Leaky Nun

in general they're all quotients of the ring?

Tobias Kildetoft

Well, given a simple $R$-module $L$, can we define a module homomorphism $R\to L$?

(non-zero one)

Leaky Nun

07:54

pick a nonzero element of $L$ lol

Tobias Kildetoft

yep

Leaky Nun

anyway this seems to be saying that every simple $\Bbb CG$-module is projective

Tobias Kildetoft

well yes, since the ring is semisimple

Leaky Nun

aha, this is saying that the $\Bbb CG$ is semisimple

so in general a ring is not semisimple

Tobias Kildetoft

very true

Leaky Nun

07:56

maybe Artin and Wedderburn have something to say about this

Tobias Kildetoft

I can guarantee that both of those people had a lot to say about this subject in their time

Leaky Nun

according to wiki, semisimple rings are all artinian

Artin–Wedderburn theorem

In algebra, the Artin–Wedderburn theorem is a classification theorem for semisimple rings and semisimple algebras. The theorem states that an (Artinian) semisimple ring R is isomorphic to a product of finitely many ni-by-ni matrix rings over division rings Di, for some integers ni, both of which are uniquely determined up to permutation of the index i. In particular, any simple left or right Artinian ring is isomorphic to an n-by-n matrix ring over a division ring D, where both n and D are uniquely determined.Also, the Artin–Wedderburn theorem says that a semisimple algebra that is finit...

and then it decomposes to matrix rings over division rings

so there's the classification

so $\Bbb Z$ already fails because $\dim_{\text{Krull}}(\Bbb Z) = 1$

robjohn

@AnindyaPrithvi sure... $\frac{x^x\,(x-3)^{x-3}}{(x-1)^{x-1}\,(x-2)^{x-2}}$

Robin Singh

08:11

I am going to start to learn Linear Algebra. I came across Gilbert Strang's course but someone here said that it is focused more on computation and I don't want that. Can you guys suggest a book?

Leaky Nun

Linear Algebra: A Geometric Approach

Robin Singh

@LeakyNun Can you give me a brief review?

Leaky Nun

I never read that book before lol

robjohn

@LeakyNun pimping Ted's book? ;-)

Leaky Nun

@RobinSingh I mean, watch 3b1b's series

it completely changed what linear algebra is to me, if you want my review

Robin Singh

08:14

@LeakyNun I did watch the full playlist

Leaky Nun

youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

@RobinSingh good

@RobinSingh then you can watch Ted's lecture series

(or his book that I linked to above)

Robin Singh

@LeakyNun I never came across those lectures. Thanks! :-)

Anindya Prithvi

09:26

@robjohn Thank you so much, I'll write it out as a question soon and ping ; Thanks again 😊

robjohn

@AnindyaPrithvi note that that does not agree with your value for $n=3$

that gives $\frac{27}4$

Anindya Prithvi

yes I have seen that

it goes 0^0

robjohn

@AnindyaPrithvi but in this case, that is $1$

Numerically, $\frac{27}4$ looks right

Anindya Prithvi

@robjohn The key says 27/16

I am attaching the file

@robjohn numerically 27/16 fits...I checked on calc

robjohn

I just took $\binom{3n}{n}^{1/n}$ numerically, and it tends to $\frac{27}4$ pretty closely

$\binom{2n}{0}=1$

Anindya Prithvi

09:34

@robjohn desmos.com/calculator/bdauauw8va

3ncn/2ncn....not the above...I think it should diverge

robjohn

It is not $\binom{2n}{n}$ in the denominator

Anindya Prithvi

Ohh so the general representation was for the num and den seperate?

5 hours ago, by Anindya Prithvi

Cool! Thankss, btw any ideas on how to generalize the above for
$$\dfrac{{xn}\choose{(x-2)n}}{{(x-1)n}\choose{(x-3)n}}$$ ....For x=3, I got 27/16...and x=4 gives 64/27 ....kind of seeing a pattern

@robjohn so this is for $({xn} \choose {(x-2)n)})^{\frac{1}{n}}$ ?

robjohn

$\left(\frac{\binom{3n}{n}}{\binom{2n}{0}}\right)^{1/n}$ for $x=3$

Anindya Prithvi

Ohhhh I see

@robjohn I interpreted it wrongly..Thanks...did you go by the attached method?

Or some known approximations?

robjohn

I just simplifed it to $\frac{\binom{xn}{n}}{\binom{(x-2)n}{n}}$ and used Stirling

Anindya Prithvi

09:43

ohh so, stirling is a sword...I thought stirling will not give me a rational answer in p/q form considering (n/e) in the expression

Thanks a lot

OOOVincentOOO

10:19

If possible: could someone give me advice how to improve my question on SE? I get only some comments and no answers. Any help is welcome.

2

Q: Is Primegap $\lim\limits_{n \to \infty} \frac{3g_{n}^2}{p_{n}}=0$ known in literature?

After analysis the following limit for prime gaps was found: $$\lim_{n \rightarrow \infty}\frac{3g_{n}^2}{p_{n}}=0$$ Is this limit known in literature? Is the presented method valid (see text below)? Closest found is: Oppermann's conjecture. Method. Two functions are defined: $\varepsilon_1$ and...

robjohn

10:39

If the limit is $0$ is the $3$ really necessary?

OOOVincentOOO

Good remark, the number 3 is what I found by correlation. The 3 and the 2 (from square) are the best fit. So according comment Erick Wong this limit is slightly stronger then existing ones (but not as strong as observed).

Balarka Sen

12:04

Are there any representation theorists out here

Edward Evans

Tobias probably

Tobias Kildetoft

I can neither confirm nor deny those allegations

Edward Evans

hahah

Balarka Sen

@SayanChattopadhyay Oh fantastic, I came here to ask the same question. I think this is because if $V$ is an simple left $\Bbb CG$-module, then it must be a module over the various matrix algebras $\Bbb CG$ breaks up into. There aren't too many simple modules over a matrix algebra.

The only nontrivial simple left $M_n(\Bbb C)$-module is $\Bbb C^n$.

Mike Miller

I really liked Leaky's prime number intuition

Balarka Sen

12:09

Artin-Wedderburn is really a noncommutative version of primary factorization

be back later

Mike Miller

You know that representations split as sums of irreducibles, and that this splitting is preserved by direct sum, and that quotients split. So since every finite dimensional G-module is finitely generated, it's a summand of some (CG)^n; and if it's irreducible it is a summand of CG itself, right?

Doesn't need AW I don't think

Tobias Kildetoft

Right, as we noticed above, it works for any ring where all simple modules are projective

Balarka Sen

13:57

@MikeMiller Yeah that's a good way to see it

+1

I realized this after I left

Tobias Kildetoft

The prime number intuition of course breaks down once the ring is not semisimple

That is when you break out the Lego block metaphor instead

Balarka Sen

Yeah, that $\Bbb CG$ is semisimple is literally Artin-Wedderburn

Tobias Kildetoft

What? Artin-Wedderburn is a statement about semisimple algebras

Balarka Sen

Eh, it's Maschke

Tobias Kildetoft

Yeah

Thorgott

14:02

You can also arrive at the conclusion of Artin-Wedderburn explicitly for $\mathbb{C}G$, without invoking the general theorem

Balarka Sen

The general theorem is very natural so I don't care

Tobias Kildetoft

Yeah, Maschke plus Schur gets you there nicely

Thorgott

I've never learned the general theorem lol

Tobias Kildetoft

Actually, Maschke plus Schur is basically the general proof, isn't it?

Balarka Sen

Yeah

The general fact is that any simple left-Artinian ring is a matrix ring over a division ring

Mike Miller

14:09

I learned the proof for quals and forgot it the day after I passed

Balarka Sen

I am trying to prove representation theory can be done without character theory

Tobias Kildetoft

Prove to whom?

Also, just because something can be done does not mean it should

Balarka Sen

I don't understand characters

I refuse to use them

Tobias Kildetoft

Then fix that

Balarka Sen

I refuse them, in short

Tobias Kildetoft

14:11

They encode some nice stuff

Balarka Sen

Deformation of characters

Thorgott

let me know when you got a proof of the odd order theorem

Balarka Sen

Oh yeah shit that shit is bizarre

That's not a proof

It's magic

Characters take algebraic values or some absolute garbage

Thorgott

tell me a representation-theoretic, but not character-theoretic proof of Burnside for starters

would be helpful if you have one, cause the character-theoretic proof is magic and the group-theoretic proofs are too complicated

Tobias Kildetoft

I think I did a decent job in my notes of making the idea of taking the trace of a representation appear natural

Balarka Sen

14:14

Trace is the unnatural part

Trace of a matrix is meaningless

Sum of diagonal entries??? What

Mike Miller

You don't believe that you like Ric

Tobias Kildetoft

@BalarkaSen Here is one road to get there: Let's take an endomorphism of an irrep (but endomorphism just as a vector space)

Thorgott

something derivative of the determinant

Tobias Kildetoft

Let's call it $\varphi$.

Mike Miller

The Ricci curvature of a representation

Balarka Sen

14:15

@MikeMiler Ric(X) measure how fast the volume of a geodesic ball bubbles in the X direction

No sum of diagonal entries

It's the 2nd order term in the Taylor expansion of vol

Tobias Kildetoft

Now, let us turn this into an endomorphism of representations by the usual Maschke way, and call the result $\widetilde{\varphi}$

Mike Miller

Yeah but that's because determinant has derivative = trace

Tobias Kildetoft

Now, we know by Schur that this is in fact just a scalar, let's call it $\lambda_{\varphi}$

So we ask: What scalar is it, and the answer is $\frac{1}{m}\mathrm{Tr}(\varphi)$ where $m$ is the dimension of the original irrep

Sayan Chattopadhyay

@MikeMiller This is neat

Mike Miller

It's Leaky's idea

Or Maschke's

I'm just repeating and rephrasing

Tobias Kildetoft

14:19

Of course, the trace here is just a special case of the general concept of a symmetrizing trace on an algebra

Mike Miller

Of course

Tobias Kildetoft

So it is a way to turn the algebra of matrices into a symmetric algebra

Sayan Chattopadhyay

@BalarkaSen Oh yeah this was my question as well, why trace?

Tobias Kildetoft

@SayanChattopadhyay See what I wrote above for one reason to consider the trace

MathStudent

Hi all

According to Taylor's theorem (or just using L'Hopital), it should hold that $\lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/x = 0$. However, when plotting $(log(1+e^x) - log(2) - x/2)/x$ or even just calculating it for small $x$, it looks like it goes to somewhere around $-0.28$. Does anyone see what's wrong here?

Mike Miller

14:27

W|A doesn't agree with you that the plot looks like it goes to -0.28

Are you sure you didn't make an arithmetic mistake

MathStudent

Weird. You're right, it doesn't. I had just typed the function in to google and that's what it gave me. I guess I can't trust google for plotting functions. Thanks.

Mike Miller

The problem is that google interprets log as base 10. If you type ln you should get the correct plot.

Balarka Sen

loogle

log base googol

mwahahah

MathStudent

Thank you! That idea might have just not only solved that problem for me but also a bigger problem. I'm guessing that the log function built into R might also be base 10 then

Mike Miller

Yeah just compute log(10) to find out

MathStudent

14:39

@BalarkaSen hahah. pretty funny. i only just found about googol recently

@MikeMiller will do, thanks!

Leaky Nun

@MikeMiller what is this, actual numbers?

Balarka Sen

lolol

Hawk

Let $N$ be normal in $G$ and $G_0$ normal in $G_1$. I want to verify that $G_0N$ is normal in $G_1N$. So I let $g_1n'g_0n(g_1n')^{-1} = g_1n'g_0nn'^{-1}g_1^{-1}$ but i m stuck at this stage.

Leaky Nun

well you want to prove that it's in $G_0 N$ right

Hawk

yes

Leaky Nun

14:45

so move $g_0$ all the way to the leftmost position

Hawk

oh because N is normal

Leaky Nun

on the way you'll change $g_0$ but keep it in $G_0$

Hawk

and because N is normal the rest of the stuff lives in N and that's it!

thanks

MathStudent

15:00

If anyone is curious, R does use exp(1) as the base for its log function after all

Is it always true that $\lim_{x \to 0} f(x) g(x) = \lim_{x \to 0} f(x) \lim_{x \to 0} g(x)$ or are there exceptions? If so, when does it not? I thought it is always true but it seems like my above function multiplied by $1/x$ goes to $1/8$, not $0$ as it would if that rule applied here

Thorgott

what you wrote is always true, but I assume you wanted to write something else

MathStudent

oh yeah i did sorry

lol

$\lim_{x \to 0} f(x) g(x) = \lim_{x \to 0} f(x) \lim_{x \to 0} g(x)$

Mike Miller

If the two expressions on the RHS exist (that is, f and g converge to some number as x -> 0; in particular neither goes off to infinity), then it is a theorem that the left side converges too, and to the expression on the RHS

MathStudent

or actually i guess just for any limit not necessarily $x \to 0$

Mike Miller

If the expressions on the RHS don't exist then you can't say anything about the left side

Thorgott

15:09

this can be extended to some cases where the limit is infinite as well

MathStudent

Thanks, @MikeMiller. But why is it then, that it seems to hold that $\lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/x = 0$ and $\lim_{x \to 0} 1/x = 0$ but \lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/(x^2) = 1/8$ (according to L'hopital and also visible when plotting it)?

Thorgott

if one of the limits is finite and non-zero and the other is pm infty, then the limit of the product is pm infty

$\lim_{x\rightarrow0}1/x$ doesn't exist

MathStudent

Oh man. AKA the answer to why it seems that way is I'm stupid. Thanks

And sorry, idk what's wrong with my brain

Thorgott

it's fine, we all make mistakes

Clarinetist

15:37

People who learned analysis from Baby Rudin: did you cover Polish spaces?

Mike Miller

What's your goal with this stuff?

Thorgott

pretty sure there's no mention of Polish spaces in Baby Rudin whatsoever

Mike Miller

I've never used Polish spaces in my life for what it's worth

Balarka Sen

polish spaces is where a bunch of stochastic shit can be done i assume

Clarinetist

@MikeMiller They're coming up in my measure theory class. Baby Rudin is the textbook for the prereq course.

Balarka Sen

15:41

nailed it

Mike Miller

That'll do ya

Clarinetist

I suspect that this prof is teaching this class primarily with the perspective of preparing students to be probability researchers. I've not found a single textbook that covers this material the way he does.

It's not what I expected, but I guess I'll either get pushed now or pushed later.

Balarka Sen

A Polish space is just something homeomorphic to a complete metric space with a countable dense subspace. Most likely you won't need a lot of properties very specifically Polish

The reason this is useful in probability is because to define continuous stochastic processes often you define em on the countable dense subspace

Eg Levy's construction of the Brownian motion

Clarinetist

@BalarkaSen Yeah, my topology background is WEAK. I recognize all of those terms as topology terms, but I don't know what most of those words mean.

Balarka Sen

And then take limits to extend it to all time (time parameter varies over your space now instead of [0, infty))

taking limits is where you need complete

Clarinetist

15:44

I will have to do some really serious studying soon

Balarka Sen

Just think of trying to define a sequence of random variables $\{W_t\}$ where $t$ varies over a space $X$, not $[0, \infty)$

who said time is linear?

Lol Mike

Mike Miller

Weak bit, deleted

Balarka Sen

weak convergence of weak bits

Clarinetist

So basically, it sounds like you need this stuff to define more general stochastic processes. I'm only familiar with discrete-time and continuous-time processes.

Balarka Sen

yeah who said time is 0, 1, 2, or [0, infty)

time can be really fucked up

thats what

all you need to remember is what Levy did, which is passing to the rationals -- define it on rational times and then extend

Ted Shifrin

15:49

Howdy, a @Balarka, @Clarinet, @MikeM

Mike Miller

Hi

Balarka Sen

rationals is a countable dense subset of R, a complete metric space

Clarinetist

I have a TON of topology to learn soon. That much is clear. This prof assumes we know about Polish spaces coming in

Morning @Ted

Ted Shifrin

I know nothing about Polish spaces. But I'm 25% Polish.

Clarinetist

Lol

Ted Shifrin

15:50

I'd do better with Russian spaces.

Clarinetist

All gibberish to me at this point minus "topological space" and "complete metric space," per Wikipedia:

"In the mathematical discipline of general topology, a Polish space is a separable completely metrizable topological space; that is, a space homeomorphic to a complete metric space that has a countable dense subset."

Thorgott

Polish is the right condition on groups to force measurable homomorphisms to be automatically continuous or something, right

Clarinetist

I think I'll have to take some days off and just dive into topology

Balarka Sen

Nobody cares about Polish group man

@TedShifrin Definition: A Polish space X is a topological space that will, over the course of three treaties and twenty years, become a disjoint union of a Prussian space, an Austrian space, and a Russian space.

(communicated to me by Fargle)

Ted Shifrin

Count on @Fargle :P

Thorgott

15:56

huh, there's an open mapping theorem for Polish groups

Clarinetist

Sometimes I wonder why stats curriculum works the way it does. You basically have three phases of stats: the stuff that only requires college algebra, the stuff that requires calculus + linear algebra, and the stuff that requires measure theory. Why not just push the math prerequisites through from the beginning and learn the advanced stuff as it should be treated?

Ted Shifrin

I think it's important to develop intuition and learn lots of examples without all the fancy machinery. This applies in virtually all areas of mathematics.

Mike Miller

Yeah the fancy stuff obscures basic understanding; you then work to see why your basic understanding matches with what the fancy stuff is doing

Clarinetist

The main problem I'm having to deal with - as a data science faculty member - is people thinking they know everything there is to know about stats when they've only taken the college-algebra-level course.

I get your points.

What I find interesting is from talking to people who hold bachelor's degrees in Stats from Canada and Europe is that they're pushed to take measure theory in their bachelor's degree. Not the case in the US. I wonder if there's a difference in priorities with countries.

Ted Shifrin

In Europe everyone taking "calculus" learns analysis, seriously.

Balarka Sen

16:00

I always thought notions of sufficiency, completeness are nontrivial in statistics.

Ted Shifrin

It's a very different approach to education.

Balarka Sen

A sufficient statistic is pretty much literally a measured foliation as Thurston thought of it

Clarinetist

Oh, that's interesting. So they dive in to things like Spivak and Apostol

Lol I have no idea what a sufficient statistic is now @BalarkaSen

Balarka Sen

Oh, I thought that's standard

Thorgott

there's no such thing as calculus here

Balarka Sen

16:01

This was in my Stat-II course.

Ted Shifrin

@Clarinet: More like Rudin than Spivak/Apostol, in fact.

Clarinetist

@TedShifrin Wow, that's intense

No idea what a measured foliation is. The standard text that I've seen people use for the master's-level sequence in statistics is Casella and Berger's Statistical Inference, which defines a sufficient statistic as given here: en.wikipedia.org/wiki/…

Ted Shifrin

Well, students select early on (like before high school?) there whether they're going to study science or other things. It's totally a different model of education than in the US.

Balarka Sen

@Clarinetist Ah I don't expect you to know what a measured foliation is, no. I was confused that you had no idea what a sufficient statistic is.

Clarinetist

Lol

Balarka Sen

16:04

Sufficiency, minimality, completeness etc are introduced in my Bachelors' course

Clarinetist

Honestly, I was disappointed in that class. The whole discussion of UMVUEs is interesting, but once you realize how restrictive the assumptions are to discuss that topic, you start wondering about the practical implications

Balarka Sen

I am very much not a stat guy, I just thought the math was interesting

Ted Shifrin

I confess that I never learned any statistics. I am not proud of this fact.

Balarka Sen

It's totally nonobvious that a boundedly complete minimal sufficient statistic is independent of an ancillary statistic.

Ted Shifrin

Balarka is speaking gobbledygook.

Balarka Sen

16:06

It has weird implications like, if $X$ and $Y$ are Gamma distributions, $X + Y$ and $X/(X + Y)$ are independent(!!)

Clarinetist

I only got into stats because I didn't appreciate how much hand-waving was done in actuarial science, and I didn't see myself going to grad school for pure math

@BalarkaSen Basu's Theorem is such a remarkable result

Balarka Sen

I agree!!

The proof is a mere 2-lines. The nontrivial part is to come up with those definitions.

Clarinetist

Completely agree

Balarka Sen

Here is what I meant by a measured foliation, by the way. So $T$ be your statistic, which is a map from the sample space $\Omega$ to $\Bbb R$. $T^{-1}(t)$ then "foliates" or partitions $\Omega$

Clarinetist

Oh, that's interesting

Balarka Sen

16:09

Ok, let's not say it that way. Let me see.

So basically the point is you have a family of distributions $X_\theta$ parametrized by $\theta$

And distribution of $T(X)$ is completely independent of $\theta$

That means you have an evolution of a measure $\mu_\theta$ on $\Omega$ such that on each leaf $T^{-1}(t)$, the evolution is constant: $\mu_\theta$ restricted to these leaves is constant.

Clarinetist

Interesting

Balarka Sen

That is somehow exactly what a measured foliation is, but just with nice smooth spaces ("manifolds").

Clarinetist

Some day, I'll get myself to open up Schervish's text lol

Balarka Sen

You have some evolution of a measure which are time independent along leaves, but varies in the "transverse" direction

Oh, I don't know that book

Clarinetist

It's an old book that people still use for PhD-stats qualifying exams in mathematical statistics

Theory of Statistics, Schervish

Balarka Sen

16:12

Ah alright

Clarinetist

So, it looks like I have a ton of topology to learn on my to-do list

I don't envy full-time teaching faculty these days who are online. It's gotta be overwhelming.

Balarka Sen

@TedShifrin Suppose you have two separate radioactive materials, and $X$ and $Y$ represent the time that an $\alpha$-particle emanates out of one of them, respectively.

These are random times, so random variables.

Clarinetist

One 3-credit class is enough for me already

Balarka Sen

The theorem is that $(X + Y)/2$ (average time) and $X/(X + Y)$ (proportion) are independent.

Which is algebraically totally nonsense ("of course those expressions have a relation, multiple one with the other, get X.. oh wait, X is random, but surely something works...")

Clarinetist

@BalarkaSen Yeah, it's hard for me to accept when someone claims two random variables are independent these days without a proof by definition. Too many people handwave that step.

Balarka Sen

16:17

Independence is such a nonobvious notion

Clarinetist

I remember I had a really nasty independence problem on MSE one time... let's see if I can find it

Here it is. math.stackexchange.com/questions/3702389/…

Balarka Sen

Ahh yes classic

You can say something intuitive about that though, multiply both with $1/\sqrt{2}$ and note that you're just rotating the 2D normal distribution

And Gaussians are rotationally symmetric

Clarinetist

My MS qualifying exam dealt with piecewise Jacobian transformations. THAT was something to study lol

user330477

18:02

Example of a function for which contour inequality is equality

geocalc33

19:29

$z^2-x^2-y^2 = 1 \implies x^TQx = 1$$where

$$Q = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$

I'm stuck on how to rotate the original function so that the forward sheet lies in the first octant (+,+,+). I'm pretty sure I have to change the entries in the matrix Q

is that right?

nevermind

Anonymous

20:41

If a sequence converges to $0$, does its rearrangement also converge to $0$?

gparyani

I'm planning to ask a question on the main site that basically asks, "given how Stack Exchange runs moderator elections, is it possible for a candidate to win if the election was being run with x seats, but lose if the election was being run with x+1 seats?". I'm not familiar with the site rules, so what sort of info am I supposed to put in it in order for it to be a good fit?

Anonymous

Or rather if a sequence does not converge to $0$, does its rearrangement also not converge to $0$?

Hawk

what's the difference between a normal extension and the algebraic closure?

Ted Shifrin

@Hawk: $\Bbb Q[i]$ is a normal extension. So is $\Bbb Q[\root4\of 2,i]$. The algebraic closure of $\Bbb Q$ is gigantic.

Right. A normal extension is a splitting field of a single polynomial.

Every polynomial splits in the algebraic closure.

Leaky Nun

@Hawk why do you think they're the same?

Hawk

20:55

@LeakyNun i read the definitions and they look like they are saying the same thing

Leaky Nun

can you tell us what definitions you read?

Hawk

it was just on wikipedia

Leaky Nun

can you copy the wikipedia definitions?

Ted Shifrin

@Hawk: What we learned from our discussion the other day is that you often don't pay attention to little details that are super important in mathematics. You need to force yourself to pay attention to them.

Sayan Chattopadhyay

On compact smooth manifolds, do vector fields always admit global flows?

Hawk

20:57

well i m reading them now and there is a requirement that we have to use irred polynomials to talk about normal extensions @LeakyNun

Leaky Nun

that's not a definition

Ted Shifrin

Yes, @Sayan.

Hawk

@LeakyNun F is alg closed if every non-constant poly in F[x] has a root in F. L/K if every irreducible poly over K that has at least 1 root in L splits over L.

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$Mathematics$ Mathematics