Mathematics - 2024-11-06

psie

00:15

How long can it take until an edit is approved (or rejected)? Maybe I didn't do the right thing, but I wanted to put a bounty on a question (this one), but I only want to bounty it if it looks nice (currently it doesn't I think). So I did edit the post somewhat, but the question is the same. Maybe I should have just asked a new question, but it is literally the same.

Ben Steffan

Depends entirely on whether people are around to review it

I wouldn't put a bounty on a question this old though

there was a question about this on meta recently

psie

yeah, I should have just asked a new question, taking the risk of getting closed as a duplicate

Ben Steffan

well the question has an answer

if you post a question, you should make sure to explicitly say why that answer isn't enough

psie

yes

Ben Steffan

oh, actually your question can't be closed as a duplicate of that one haha

you can only close as duplicate if the target question has an upvoted or accepted answer

but the answer there has no upvotes

psie

00:21

ah, interesting

Ben Steffan

so I think this is a good case

psie

ok, well, now I just have to wait until my edit is reviewed :( unless there's a way to cancel the edit, which I'd happily do

Ben Steffan

I think the edit is a bit too much

it's essentially a complete rewriting of the question

psie

yeah

psie

00:50

when you gave your talk @BenSteffan, did you use beamer in LaTex? :)

Ben Steffan

no, we generally do blackboard-only talks

psie

oh, ok

Ben Steffan

unless it's a zoom talk

psie

always wanted to practice my skills in beamer

it's not easy

if you're a beginner

Ben Steffan

you get the hang of it relatively quickly I think

but it has its quirks

Thorgott

00:52

I've done a talk with slides over zoom once, it's eh

Ben Steffan

also depends on how good your general latex skills are of course

I've done two

Thorgott

all the other talks I've done over Zoom, I just drew with a tablet

Ben Steffan

yeah, if I have to give a talk like that again I would probably do the same

beamer talks are difficult

you'll inevitably be too fast

(for people to follow)

Thorgott

but nothing beats the primal joy of getting your hands dirty with some chalk

Ben Steffan

indeed

psie

00:56

I found it extremely challenging to get pictures where I wanted them in my beamer presentation...there was a point where I became so frustrated that I thought about doing everything in google slides and use their add-on tool to get equations as pictures into the presentation...at least there I knew how to get the pictures where I wanted them

Ben Steffan

I actually always manage to cover myself completely in chalk

I also have a bad habit of leaning on the blackboard when I'm not writing

Thorgott

the classy spots of dried chalkwater on one's shoes

Ben Steffan

mathematicians are a weird bunch

they built new lecture halls here a few years ago, for shared use by different faculties

but they only installed whiteboards

Sine of the Time

-_-

Ben Steffan

our department was very unhappy with this after the first few math lectures people had to give there, so they asked the uni if they couldn't swap out the whiteboards for blackboards

response: "sure, if you pay for it"

so now there's blackboards

we don't even use these lecture halls all that much. it's mostly the first year bachelor students that get to see them from the inside, mainly because they're huge

Derso

01:57

Hi, guys! Can anyone check my answer to this question? math.stackexchange.com/questions/4994416/…

Thomas Finley

04:56

While studying about PDEs I found that we can calculate the general solution of a PDE from it's complete solution.

The method goes like this:

If say, a PDE $F(x,y,z,p,q)=0$ where $p=z_x,q=z_y$ is given, and $f(x,y,z,a,b)=0$ is a complete solution of $F,$ then assume $b$ to be an unknown function of $a$ say, $b=\phi(a).$

So, the complete solution $f$ becomes, $f(x,y,z,a,\phi(a))=0.... (1)$

Now, differentiating the above equation wrt $a$ we get, $f_a(x,y,z,a,\phi(a))+f_b(x,y,z,a,\phi(a))\phi'(a)=0.... (2)$

Soham Saha

06:16

Given answer says (D), but if we try $r=0$, shouldn’t the correct option be (A)? Am I missing something obvious, or is the answer given wrong?

Soumik Mukherjee

Well if D is correct then A will also be correct

Soham Saha

@SoumikMukherjee And (D) is correct by AM-GM I think

Soumik Mukherjee

D is not correct as you already pointed out

Soham Saha

Yeah I mean (D) covers all cases for non-zero r

Soumik Mukherjee

Woke up and found out that Trump is winning -_-

Sine of the Time

06:30

I've slept 3h

Soumik Mukherjee

Why?

@SineoftheTime what are you currently studying ?

Sine of the Time

Nothing, I gave an exam and I'm taking a couple of days of rest

Soumik Mukherjee

Taking rest by only sleeping for 3h? :"(

Sine of the Time

After the exam, I was tired and slept 1h between 8 and 9 pm

Then I woke up and I couldn't go to sleep until 4.30 am

Soumik Mukherjee

07:16

Oh

Pizza

09:23

Hi

Soumik Mukherjee

Hi

think_meaning_builds

09:48

Hi

D S

10:13

I'm a bit confused with notations in sets. Suppose $S \subset \Bbb R^3$, then $S$ has elements of the form $(a,b,c)$, where $a,b,c$ are reals. So, is $S \times S =S^2$ a subset of $\Bbb R^6$? That would mean elements of $S^2$ are of the form $(a,b,c,d,e,f)$. But to me, it seems that the elements should be of the form $((a,b,c),(d,e,f))$. Or are the two equivalent?

Soumik Mukherjee

@DS Yes, and they are equivalent

D S

@SoumikMukherjee thanks, that would really help me to shorten a proof.

Soumik Mukherjee

np

The only thing you need to mention is that $(a,b,c),(d,e,f)\in S$

Jakobian

12:13

@DS In mathematics we identify the tuple $((a_1, ..., a_{k_1}), (a_{k_1+1}, ..., a_{k_2}), ..., (a_{k_N+1}, ..., a_m))$ with the tuple $(a_1, ..., a_m)$

Are they literally the same? No they aren't

so you would identify in this way the sets $\mathbb{R}^6$ and $\mathbb{R}^3\times \mathbb{R}^3$ even though they aren't literally the same

psie

Does anyone know if this theorem remains the same when the measure spaces are complete? I know Tonelli-Fubini has its own version when the measure spaces are complete, hence my doubt about whether or not this theorem also needs to be altered if the measure spaces are complete.

Jakobian

@psie what do you mean by that.

Of course this theorem holds when $\mu, \nu$ are complete as long as they are $\sigma$-finite. The only thing changing that I can think of is if we replace $\mu\times \nu$ by its completion

Fubini's theorem

In mathematical analysis, Fubini's theorem characterizes the conditions under which it is possible to compute a double integral by using an iterated integral. It was introduced by Guido Fubini in 1907. The theorem states that if a function is Lebesgue integrable on a rectangle X × Y {\displaystyle X\times Y} , then one can evaluate the double integral as an iterated integral: ∬ X × Y f ( x ,...

psie

@Jakobian yeah, if we have $\overline{\mu\times\nu}$, I don't think we get that $\overline{\mu\times\nu}(E)$ can be written as integrals with respect to $d\overline{\mu}$ or $d\overline{\nu}$, right?

Jakobian

12:39

Here Bogachev by $\mu\otimes \nu$ denotes the product measure on the completion of the $\sigma$-algebras

but $\mathcal{A}\otimes \mathcal{B}$ doesn't denote the completion, let me think

This is actually the completion of this $\sigma$-algebra so yes this is actually what you are searching for @psie

psie

ok 👍

Jakobian

But it does seem necessary for $E_x$ and $E^y$ to not be always measurable, but for a.e. $x$ and a.e. $y$

nonetheless the functions should be a.e. well-defined so we can still talk about their measurability

nickbros123

12:56

Is the following true in a general setting? $[G:H\cap K]=[G:H][H: H \cap K]$ where ofcourse $K,H \leq G$ and G is a group(possibly infinite,)

Here [G:K] is the number of cosets of K coming from H

When all these things are finite, of course

Jakobian

@nickbros123 should be true when all those things are finite or infinite

No need for $H\cap K$ just assume $K\subseteq H$

But yes. This always holds

Thorgott

it's even true if you interpret those quantities as cardinals, but you typically don't need to worry about infinites more precisely

nickbros123

13:27

Yeah got it now, thanks

I just had to look at $G=\cup_{i=1}^{[G:H]}g_i H $ and write $H=\cup_{j=1}^{[H:K]}h_j K$ and notice that $g_{i} (h_{j}(K))=(g_i)(h_j) K$

this concludes the main step in trying to prove that if $[G:H]=m$ and $[G:K]=n$ then $lcm(m,n) \leq [G:H \cap K]\leq m \cdot n$

Jakobian

13:44

yeah, I meant what Thorgott just said

except the remark starting with "but" of course

Thorgott

:P

Jakobian

14:20

@Thorgott Do you think there is some kind of equivalent formulation of being a totally disconnected compact Hausdorff space in terms of surjections into $\{0, 1\}^\kappa$ for some cardinal $\kappa$?

I just found this property of the Cantor cube

> AE(0), an absolute extensor for compact zero-dimensional spaces. (Every map from a closed subset of such a space into a Cantor cube extends to the whole space.)

which may be useful to me

I want, given a surjection $f:S_0\to [0, 1]$ where $S_0$ is a zero-dimensional compact space, for there to exists a lift $f_0$ with $f = q\circ f_0$ where $q:\{0, 1\}^\omega\to [0, 1]$ is the canonical quotient map

Do you think that's possible

and I guess $f_0$ a surjection

At least I THINK that's what I want

never mind, let me think about it

Thorgott

14:48

is zero-dimensional compact Hausdorff the same as totally disconnected compact Hausdorff? (i.e. profinite spaces)

Jakobian

@Thorgott yes

for compact Hausdorff spaces, totally disconnected and strongly zero-dimensional are equivalent

nickbros123

15:02

I was once trying to prove a theorem regarding homomorphisms on groups, and I was using V and W as notation for groups. I couldnt prove it till I replaced them with G and H lol

Jakobian

Baire space (set theory)

In set theory, the Baire space is the set of all infinite sequences of natural numbers with a certain topology, called the product topology. This space is commonly used in descriptive set theory, to the extent that its elements are often called "reals". It is denoted by NN, or ωω, or by the symbol N {\displaystyle {\mathcal {N}}} or sometimes by ωω (not to be confused with the countable ordinal obtained by ordinal exponentiation). The Baire space is defined to be the Cartesian product of countably infinitely...

this wikipedia page is stupid - concept in set theory? Concept in topology?

Both are concepts in topology

someone who was editing this didn't know what they were talking about

like, sure, The Baire space is used in set theory as well, but its still an important concept in topology

Jakobian

15:25

I feel like if we remove from a compact connected Hausdorff space $Y$ a zero-dimensional set $X$ then $Y\setminus X$ should contain the Cantor set

ah no that's not true

Because one can take $Y$ to be $[0, 1]$ and $X$ to be irrational numbers in there

which shows that a zero-dimensional space can have a $> 0$ dimensional compactification for reasons other than part of remainder surjecting onto $[0, 1]$

I have no idea what $\beta (\omega^\omega)$ is

Jakobian

16:00

My discussion inspired me a bit. Still, not sure if what I have is enough to solve this problem I am struggling with, probably not

psie

16:32

Suppose $m$ is complete and $f$ is a measurable map that pushes $m$ forward. Will the pushforward measure be complete? My guess is not unless the preimage of $f$ yields a null set for every null set, is that right? I don't know how to phrase that correctly.

I googled quite some time for this, but couldn't find it on the main site.

The reason I am asking is because I have a pushforward measure $m_\ast$ that is the product of two measures $\rho$ and $\sigma$, so $m_\ast=\rho\times\sigma$. If I complete it, will $\overline{\rho\times\sigma}$ still be a pushforward measure somehow?

In other words, can we obtain the completion of $m_\ast(E)=m(f^{-1}(E))$ by replacing $m$ with its completion?

Ryder Rude

16:48

hi

Ben Steffan

hi

Derso

hi

Jakobian

17:39

1

A: Spaces with every compactification $0$-dimensional which aren't locally compact

The Sorgenfrey line is an example, see The Sorgenfrey line has no connected compactification, by Emeryk and Kulpa.

I got my question answered, but sadly I think its wrong

I mean I don't just think that, it is wrong

having connected compactification and having a compactification which isn't totally disconnected are two different things, and I can prove Sorgenfrey line has a compactification which isn't totally disconnected

Sine of the Time

19:23

:/

pie

19:34

How many ours do you study math per day in average?

Ben Steffan

about 40

pie

I heard that Erdos studied 19 per day

Ben Steffan

wonder how he did that :)

realistically I think I engage with mathematics for about 10-12h every day right now

but I'm really bad with times and dates

pie

I also heard that Richard Borcherds used to study 16 hours a day. This makes me feel like all my efforts in studying math aren't enough.

leslie townes

erdos was on pills, too. maybe not a role model.

Ben Steffan

19:38

that's what I was getting at

every time people bring up erdős I flinch a little

leslie townes

i was surfacing the comment for someone who might not have gotten the joke :)

Ben Steffan

ok ok :)

pie

How many hours do you think top mathematicians used to study?

Ben Steffan

maybe 1 or 2

pie

Euler for example

leslie townes

19:40

pie, i never know what to make of reports like that, people vary so much in what they mean to 'study' for x hours, if they are even self reporting that or just having stories told about them.

suffice it to say that even most accomplished career mathematicians are not going to have borcherds's career

Ben Steffan

comparing yourself to other people is very unhelpful here

leslie townes

which maybe you see as discouraging, but it is also encouraging. lots of stuff to do in math without being 'the next' somebody

pie

I am a self learner so I don't know what is the norm, is 7 hours a day good or bad or too much ? the only information I can easily find by googling is only for top mathematicians.

Ben Steffan

this is unanswerable

how much progress do you make? what are your goals? can you bear the workload? how's your work-life balance? etc.

leslie townes

it isn't a task like plowing a field, where more hours spent doing it translates into some concrete amount of accomplishment, either. a lot of people (even famous ones) will say that you need to make allowance for "downtime" and that you might even get more productive thoughts during or as a result of downtime, than during or as a result of being "on the clock"

a lot of people who do math find themselves just thinking about it and reading things for recreation even if it isn't "study" in the sense that preparing for something like an exam is study

pie

19:45

@BenSteffan Even I don't have a clear answer for these questions.

Ben Steffan

...and thus the question you asked cannot be answered :)

pie

I just like math I want to learn the biggest amount of it, this is the only reason

I wish I can have any career in PURE math :)
But things don’t always go as planned.

psie

Let $f$ be nonnegative or integrable such that $f(x)=g(|x|)$ for some function $g$ on $(0,\infty)$, and suppose $$\int f(x)\,dx=\sigma(S^{n-1})\int_0^\infty g(r)r^{n-1}\,dr\tag1$$holds. It is claimed that if $|f(x)|\leq C|x|^{-a}$ on $B=\{x\in\mathbb R^n:|x|<c\}$, then $f\in L^1(B)$. Here's my attempt:$$\int_B |f(x)|\,dx\leq\int_B C|x|^{-a}\,dx\leq\int C|x|^{-a}\chi_B(x)\,dx$$but can we write $C|x|^{-a}\chi_B(x)$ as some function $g(|x|)$ in order to apply $(1)$?

Above, $C,c>0$.

Also, $a<n$.

leslie townes

psie: chi_B(x) depends only on |x|, if that helps?

psie

It depends on x, or?

Ben Steffan

20:00

I think it's pretty funny that this question has not received an answer mathoverflow.net/questions/481662/…

leslie townes

psie: think of what chi_B actually is. in symbols you have something like chi_B(x) = chi_{[0,c)} (|x|) but one should maybe not get too distracted by symbols here

psie

true, there is some dependence on |x| there...hmm

that was kind of concealed in all the other symbols :) but you saw it

thank you

leslie townes

if ted were here he'd ask you what "geometric" fact about a shape B will make chi_B(x) expressible as a function of |x|

theres a mental picture here

psie

haha, that would be Ted's line indeed

Thorgott

@BenSteffan lol

nickbros123

20:13

@pie 6 throughout the semester, ~8 in holidays, between 0-4 during finals week

2

copper.hat

i'm sure Ted is off celebrating the results

Jakobian

20:35

@pie 1 or 2 hours probably

@nickbros123 in total

I guess I'm not a good learner

But I got the spirit

Pizza

21:00

Hi

Sine of the Time

@Pizza hello

AlexanderPope

21:26

Hi, silly question I am embarrassed I cannot answer! Given $p$ in the unit simplex, is it true that $\{p + \lambda(q-p): q \in unit simplex, \lambda \geq 0\}$ is closed?

Dannyu NDos

The unit simplex is compact, so yes.

Thorgott

what is the "unit simplex"

leslie townes

if p is in the interior that will be R^n (or whatever space this is living in), so yes. if p is on the boundary it feels like you should get some finite intersection of half spaces, so still yes. i assume the "unit simplex" is at least a polytope or something in R^n.

AlexanderPope

\{p \in \mathbb{R}^n: p_i \geq 0, \sum_{i}p_i = 1\} is the definition of unit simplex

Dannyu NDos

Nvm, I guess $\lambda$ ruins compactness.

AlexanderPope

21:31

I tried just using a basic sequential argument but couldn't prove it. Maybe I am missing something. I was reading a paper on fixed point theorems, and this question arose.

leslie townes

i would try to think of some characterization of the unit simplex as an intersection of half spaces, so that you can describe the 'directions' involved in {lambda (q - p): lambda >= 0, q in the unit simplex} similarly. of course translating the thing by p is not going to affect whether it is closed.

as ted might say, it might help to think geometrically to think about a proof argument, just attacking it with sequences seems tricky as the representation of a point in that set in terms of q and lambda will in general not be unique and it feels like you would need to work 'choice' representations to push that argument through.

geometrically it feels to me that thing is oging to look like the simplex with one or more of its walls blown out to infinity, but (if you didn't start with an interior point) at least one and maybe more than one of them still there. my mental vision of this suggests it is closed.

you can tell i'm not a geometer because when i talk about this stuff i sound like a hallucinating crazy person

Ben Steffan

well it's certainly true, but I don't immediately see how to prove it

:^)

You can draw a picture for $p = 0$ and $n = 1, 2$ to convince yourself of this

leslie townes

i don't see what's specific about the unit simplex here, it feels like any convex polytope would do. maybe even any polytope

i don't know how helpful that it is, but it seems worth remarking upon

Ben Steffan

yeah

AlexanderPope

Ok, thank you, everyone, I think it is true too! Maybe I'll ask this as a question if I don't see how to prove it in the next hour or so!

Sine of the Time

21:41

@BenSteffan chat is too small to contain the proof

ILikeMathematics

Hi

Ben Steffan

hmm, maybe something crude like the following works: Consider the maps $\phi_k\colon \Delta^n \times [k, k + 1] \to \mathbb{R}^{n + 1}$ with $\phi_k(x, t) = tx$ for all $k \geq 0$. Then your set (for $p = 0$) is given as the union of the images of all these $\phi_k$, and each image is compact, so if you have a sequence that doesn't converge in the set, then it eventually will have to escape each of the images (this requires some arguing about the intersections) and therefore go to infinity

leslie townes

huh. maybe even any closed convex subset of R^n?

Ben Steffan

Hi

Jakobian

for a second I thought you are responding to leslie and I was like, "dementia?"

ILikeMathematics

21:49

Any ideas on proving that for any irrational $\tau$, for every $\varepsilon > 0$, there exist $n, m \in \mathbb Z$ with $0 < m \tau + n < \varepsilon$? I'm not really sure in what way it is true, if we consider $\pi$ and set e.g. $\varepsilon = 0.1$, then what could $m, n$ be?

Ben Steffan

@Jakobian in my current state that would not be entirely unlikely haha

@ILikeMathematics $n$ something very negative, $m$ something roughly $|n| / \tau$

Jakobian

@ILikeMathematics since $\{m\cdot \tau\}$ is a sequence uniformly distributed mod $1$, all you have to do is find $m$ for which its between $0$ and $\varepsilon$

then let $n$ be the negative of the floor of $m\cdot \tau$

ILikeMathematics

@BenSteffan Oh

Ben Steffan

but yeah, what Jakobian is saying

leslie townes

@Jakobian Hi

ILikeMathematics

21:54

@Jakobian What's 'uniformly distributed'? Haven't covered that yet

Jakobian

It means that if you count the number of $a_k$ in any given fixed interval $(a, b)$ of $[0, 1]$ for $k\leq n$, divide it by $n$, then take the limit $n\to\infty$ then the limit will be $b-a$

in particular it implies this is a dense sequence in $[0, 1]$

Thorgott

@leslietownes if you take the closed unit ball in $\mathbb{R}^2$ and $p=(-1,0)$, then the space you get is the union of $\{p\}$ and the open right half-plane, so not closed

probably any polyhedron does the job, but that could be annoying to prove

Ben Steffan

hm, maybe it was foolish to assume the choice of $p$ doesn't matter

although the standard simplex lives in a single quadrant/octant/... so

Thorgott

@AlexanderPope so if $p$ is in the interior, then this will be the entire hyperplane containing the unit simplex, if $p$ is in one of the open faces, it will be a half-space in that hyperplane, if $p$ is in the intersection of exactly two faces, it will be the intersection of two such half-spaces, etc.

that should be the geometric picture

ILikeMathematics

@Jakobian Thanks

Well I'll have to do this without all that terminology, but I guess that'll be possible with just the idea

Ben Steffan

22:04

the density statement you should be able to prove on your own

of course if you don't know what 'dense' means you just prove that the sequence gets arbitrarily close to any point mod 1

ILikeMathematics

Thanks.

(By the way: for some reason the ChatJax rendering doesn't work anymore, javascript: doesn't execute in the address bar...)

Ben Steffan

works fine for me

Claudio

@ILikeMathematics are u using the extension for chrome?

leslie townes

23:09

@Thorgott oh good point

thanks

Jakobian

23:24

@leslietownes hi

robjohn

@ILikeMathematics It works for me. What OS and browser are you using?

psie

@copper.hat sorry for the ping, but do you happen to know about pushforward measures? I'm stuck with this problem of extending a theorem in Folland's for Lebesgue measurable functions. He says to consider the completion of $\rho\times\sigma=m_\ast$, where $m_\ast$ is a pushforward measure of Lebesgue measure under a homeomorphism $\Phi$. I am stuck at figuring out what the completion of $m_\ast$ is. Is it also a pushforward measure?

$m_\ast$ also happens to be the product of two measures $\rho$ and $\sigma$, hence why I wrote $\rho\times\sigma=m_\ast$. But $m_\ast(E)=m(\Phi^{-1}(E))$ where $m$ is Lebesgue measure.

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