Mathematics - 2024-12-03

leslie townes

02:12

"it is used in several research papers, indicating that it is a common phrase" ... it's nice when bogus reasoning is that transparent

haha

Xander Henderson

Did you not read the comment?! It said SEVERAL PAPERS!

SEVERAL!

MEANING LOTS AND LOTS AND LOTS!

IT'S SUPER COMMON!

leslie townes

several research papers

math research being a place where people are notably very sensitive to common English usage and extremely careful about word choice and phrasing

Xander Henderson

You're just saying what I'm saying!

We agree!

100%!

leslie townes

yep

Xander Henderson

💯

Who's selling houses for $100k?!

leslie townes

02:26

god haha where to begin with that "word problem"

i guess the price. the price is where to begin

then there's the rest of it

chef's kiss

if we want another 2008 crisis, this is how we get it

Xander Henderson

Indeed.

In other news, Zillow thinks I only have one bathroom.

leslie townes

any room is a bathroom. i decide how i use my rooms

Xander Henderson

@leslietownes Ha!

Zillow also thinks that my house has appreciated by \$35k in the last two years. I don't buy that, particularly considering that the house across the street has been sitting on the market for \$275k for over a year.

leslie townes

04:09

sounds like you're selling houses for 100k tbqh

around here an approximate formula for the 'zestimate' is [last selling price of a nearby home] + 20%

Ray

05:53

I had a basic doubt regarding the precise formulation of simultaneous substitution in predicate logic. Let $t_1,\ldots,t_n$ be a sequence of terms and $x_1,\ldots,x_n$ is a sequence of variables. Suppose also that the free variables of $\varphi$ are among $x_1,\ldots,x_n$, and $\varphi[t_1/x_1,\ldots, t_n/x_n]$ denote the result of substituting $t_i$'s in place of $x_i$'s. Does anyone know any simple recursive definition of $\varphi[t_1/x_1,\ldots, t_n/x_n]$?

nickbros123

06:10

@XanderHenderson whats wrong?

if I say f bijects A to B, do you think of f as a bijection from A to B or do you think of, idk, f is a rabbit or something?

copper.hat

06:30

did you just pull that out of a hat?

nickbros123

yes, copper hat.

copper.hat

almost makes me want to scroll back to see what that was about :-)

Daniel Donnelly

07:57

AC/DC - Highway to Hell (Official Video)

►

Ray

08:26

Just in case, anyone in interested in a precise formulation of my earlier comment.

0

Q: On the Definition of Simultaneous Substitution of Terms in FOL formula

Consider a firsr-order language $L$. Let $\varphi$ be a formula, $t$ a term and $x$ is a variable. We define the formula $\varphi[t/x]$, which denotes the formula obtained by substituting $t$ in place of $x$, as follows (here, $P$ is a $n$-ary predicate symbol and $\dagger$ denoted undefined; it ...

maths student

10:03

Define matrix ( B ) as the transpose of the transpose of the transpose of matrix ( A^T ). Then B is A^T or A
I think it is A

Martin Sleziak

In linear algebra chatroom: chat.stackexchange.com/transcript/13473/2024/12/3

psie

11:03

I'm reading about the following proposition:

Proposition If $F\in NBV$, then $F$ is absolutely continuous iff $\mu_F\ll m$.

In the $\implies$ direction, we suppose that $m(E)=0$ for a Borel set $E$ and if $\delta$ and $\epsilon$ are as in the definition of absolute continuity of $F$, we can find open sets $U_1\supset U_2\supset\cdots\supset E$ such that $m(U_1)<\delta$ (by regularity of $m$). This kind of makes sense. If $m(E)=\inf\{m(U):U\supset E,U\text{ open}\}$, then by definition of infimum we can find such a sequence. However, it is also claimed that $\mu_F(U_j)\to \mu_F(E)$. Why is this true?

psie

11:50

It is true that $\mu_F$ is regular, but this means that the positive measure $|\mu_F|$ is regular. So $|\mu_F|(K_j)\to|\mu_F|(E)$ for some sequence $\{K_j\}$ of open sets. It is also not clear that we can simply choose $K_j=U_j$ (and why we'd get convergence for $\mu_F$ rather than $|\mu_F|$).

Pizza

12:03

hi

Soumik Mukherjee

Hi

Pizza

How is it going?

Soumik Mukherjee

13:31

@Pizza Fine, what about you?

ModularMindset

14:26

Has anyone here ever run into $$A(x)=B(x)+C(x)$$

and more to my point, has it served you any use?

To me we can look at this as a possible orthogonal decomposition of $A(x)$ into its "orthogonal" component functions $B(x)$ and $C(x)$.

Jakobian

@ModularMindset into what?

@ModularMindset no...? How is there even anything related to orthogonality. Its just a sum of functions

Thorgott

yes, I've added two functions before

Jakobian

I've removed like 3 things from my "to check" list just recently. Amazing

2 of those were just because I learned a little more about monotone normality

now that I know fully normal implies strongly collectionwise normal, and monotonically normal implies hereditarily collectionwise normal, I wonder what are relations between all those spaces

e.g. $\omega_1\subseteq \omega_1+1$ is not fully normal, so full normality is not hereditary for sure

so I guess, lets start with this, do there exists a monotonically normal $T_1$ space which is not fully normal (or equivalently, paracompact)?

Frieren

14:48

While talking about the fact that the zero product property does not hold for the canonical product of a matrix and a vector, I was asking myself: which is the correct way of phrasing this when talking about where it does not hold? More explicitly: given $n\in\mathbb{N}\setminus\{0,1\}$, a field $\mathbb{K}$, a vector space $V$ on $\mathbb{K}$ and a matrix $A \in \text{Mat}_{n \times n}(\mathbb{K})$, should I say that the zero product property does not hold on $\text{Mat}_{n \times n}(\mathbb{K}) \times V$? Or just that it does not hold on $\text{Mat}_{n \times n}(\mathbb{K})$?

Jakobian

@Jakobian correction, monotonically normal $T_1$ implies hereditarily collectionwise normal. I don't know if we can drop $T_1$ assumption but most likely not - this seems to be essential for monotone normality. It gives us various equivalent definitions from which you can then prove e.g. that the property is hereditary

ModularMindset

@Jakobian @Thorgott I mean this as a decomposition of $A(x)$ into components $B,C$ as a way to be able to work with $A(x)$ in a simpler form

I guess I'm looking for fun but not trivial examples of this

I prefer examples in analysis

Jakobian

There is an example by Cohen of a monotonically normal space which isn't strongly collectionwise normal, see here (although the author of the answer has some trouble fixing some arguments..hm)

okay so maybe there exists a hereditarily paracompact $T_2$ space which isn't monotonically normal?

maybe there already is such example

Lucky Chouhan

Is it common to feel overwhelmed and frustrated by studying math daily but not discovering anything new our own?

Jakobian

a Lindelof $T_3$ space is paracompact, so this example is a hereditarily paracompact $T_2$ space that's not monotonically normal

apparently the example is by Heath (the article seems to not be online) and there's another example by van Douwen, from an article that might not even exist

ModularMindset

15:12

I concocted a problem (well-posed btw) that no mathematician can solve

I don't think that I can even solve it

Pizza

@SoumikMukherjee I am fine thank you

ModularMindset

A,B,C conjecture

A(x)=B(x)+C(x)

Sine of the Time

skull emoji

Pizza

hi @SineoftheTime

Sine of the Time

@Pizza hi

ModularMindset

15:18

skull conjecture

Skull Conjecture: Let $X=[0,1]^n$. For all $n>2$ $X$ admits a unique codimension one surface of revolution, $L$, with a complete metric (away from the cone points) and an embedding $e :L\hookrightarrow X$, which maximizes volume while retaining constant positive sectional curvature.

Assume the cone points $p,q$ satisfy $\mathrm{sup~dist}_n(p,q)=\sqrt{n}$ where $p,q \in L$.

$L$ is referred to as the Skull

Sine of the Time

convex skull

ModularMindset

even better name lol

The Skull conjecture is true in dim. $n=3$

@SineoftheTime but convexity is not a given here in higher dimensions

Pizza

@SineoftheTime what's up?

ModularMindset

unless you know something I don't

Sine of the Time

@Pizza I'm good, what about you?

@ModularMindset it was a pun

Pizza

15:29

good thanks, I'm currently taking a look at the pdf you told me about Fourier transforms

Sine of the Time

is it useful?

Pizza

Yes

@SineoftheTime but do you happen to know on wolfram how the fourier transform is calculated?

Sine of the Time

wolframalpha.com/input?i=Fourier+transform+calculator

Pizza

but ω > 0

Sine of the Time

why?

Pizza

15:36

it is written in the text

Sine of the Time

can you send a screenshot?

Jakobian

2

A: Countable Regular Spaces Which Are Not Monotonically Normal

Completely revised. In The cometrizability of generalized metric spaces, Section $4$, Taras Banakh and Yaryna Stelmakh construct a regular topology $\tau$ of weight $\omega_1$ on $\Bbb Q$ such that $\langle\Bbb Q,\tau\rangle$ is not cometrizable and hence not stratifiable. (I have not yet gone th...

Sine of the Time

or the number of exercise

Pizza

Calculate the Fourier transform for $\omega > 0$ of the function
$f(x) = \frac{1}{x^3 - 8i}$

@SineoftheTime this is not in the pdf

Is it possible to somehow see the result on wolfram?

Sine of the Time

you can restrict the FT to the interval $[0,+\infty[$

but I'm not sure I understand why we need w>0

Pizza

15:44

I don't understand why I get that result

Sine of the Time

strange, if $\omega>0$ then $\theta(-\omega)=0$

Pizza

mmm

Soumik Mukherjee

@LuckyChouhan discovering?

Pizza

@SineoftheTime maybe to ensure the convergence of the integral and be able to apply the residue theorem

Sine of the Time

how is FT defined?

Pizza

15:57

$x^3 - 8i = (x - 2i)(x^2 + 2ix - 4)$. This identifies a simple pole at $x = 2i$ and two complex poles that do not belong to the upper half-plane, so are not included in the boundary.

i think

@SineoftheTime $F[f(x)] = \int_{-\infty}^{\infty} f(x) e^{-i \omega x} \, dx$

psie

@psie I'm still stuck on this. If anyone has any thoughts about it, let me know.

Sine of the Time

yeah maybe this has something to do with convergence

Pizza

i think this means that the integration contour is closed in the upper half-plane to ensure the convergence of the integral

Lucky Chouhan

16:18

@SoumikMukherjee I don't exactly know what to say. I'm lost.

Xander Henderson

Ugh... 20 minutes into a 3 hour final, and people are still trickling in... why can't you be on time?!

We're even all in the same time zone now!

Ben Steffan

@XanderHenderson they... they're allowed to do that?!

Xander Henderson

@BenSteffan Given the population of students I work with, yes.

They will just have significantly less time to complete the assessment.

Sine of the Time

cam you send a copy of the exam when it's over?

Xander Henderson

@SineoftheTime Why?

Sine of the Time

16:24

just to see the level

Xander Henderson

It is, essentially, high school algebra.

Sine of the Time

ok, never mind then

Binky

Hi

Sine of the Time

@Binky hi

Binky

If a chicken and a half lay an egg and a half in a day and a half, how many eggs will a hundred chickens lay in a hundred days?

Xander Henderson

16:43

@SineoftheTime I'll give you one exam problem:

Sine of the Time

@XanderHenderson thanks :)

Jakobian

16:58

@psie Can't we write $F$ as a linear combination of non-negative NBV functions, thus decomposing $\mu_F$ as a linear combination of non-negative (finite?) measures?

psie

@Jakobian you're right, we can write $F=F_1^+-F_1^-+i(F_2^+-F_2^-)$ where $F_j^\pm\in NBV$ and so $\mu_F=\mu_1^+-\mu_1^-+i(\mu_2^+-\mu_2^-)$.

I'm not sure about the nonnegativity.

But yeah, I think it's possible though.

Because $F_j^\pm(-\infty)=0$ and they are increasing.

@Jakobian So you're saying that we can apply regularity to all $\mu_j^\pm$ to obtain open sets $K_j$ such that $\mu_F(K_j)\to\mu_F(E)$. But I still fail to see why the sets we get can be chosen to be equal to $U_j$. Recall, $U_j$ are sets that satisfy $m(U_j)<\delta$.

Jakobian

@psie No that's not what I'm saying

its not regularity that you need but continuity of measure

Soumik Mukherjee

Can someone explain how they're getting the implication 'so it may be regarded as a vector space over k' from the previous part.

I guess it's something trivial but I can't see it

psie

17:15

@Jakobian ok, makes sense 👍 I guess what I was doubting is $\bigcap_1^\infty U_j=E$, but I think this is just trivial to prove.

Jakobian

@psie this is not even true

Thorgott

@SoumikMukherjee if an $R$-module $M$ is annihilated by the ideal $I$, then it becomes an $R/I$-module via $(r+I).m=r.m$ for all $r\in R$ and $m\in M$

Soumik Mukherjee

@Thorgott Oh okay thanks

psie

@Jakobian the $U_j$'s can be constructed like this; $m(U_j)<m(E)+1/j=1/j$ by definition of $m(E)=\inf\{m(U):U\supset E,U\text{ open}\}=0$. Yeah, I'm not sure $\bigcap_1^\infty U_j=E$, maybe not?

Jakobian

@psie there are Borel sets which aren't $G_\delta$

One example would be the rational numbers

psie

17:32

@Jakobian but how do you want to use continuity of measure then? We want $\mu_F(U_j)\to\mu_F(E)$. This means $\lim_{j\to\infty} \mu_F(U_j)=\mu_F(E)$, and $\lim_{j\to\infty} \mu_F(U_j)$ means either $\mu_F(\bigcap_1^\infty U_j)$ or $\mu_F(\bigcup_1^\infty U_j)$, so I'd expect $E=\bigcap_1^\infty U_j$ or $E=\bigcup_1^\infty U_j$.

I guess $\bigcap_j U_j$ (or $\bigcup_j U_j$) and $E$ can have same $\mu_F$-measure but need not equal?

Jakobian

18:10

@psie yeah I don't know. Not my field to be honest

ModularMindset

18:36

Is networking also difficult in the math world for you?

Jakobian

What is the statistics on mathematicians in academia and religion?

Scientists generally speaking have low number of people who are religious. I am wondering how it is with math

ModularMindset

there is definitely limited data

Jakobian

@ModularMindset what exactly do you mean by that

ModularMindset

but from some of the studies i looked at

tough to tell - I would predict mathematicians are not far off from the data collected about the scientists in general

Jakobian

could you link me some study that discusses actual percentages

ModularMindset

18:44

would be cool if there was some ordering Mathematicians<physicists<chemists<biologists<sports scientists

link.springer.com/referenceworkentry/10.1007/…

That link is not really study

but I honestly can't find much

there are a bunch of high profile mathematician anecodotes - but that's not terribly good I guess

Someone should do a study on this :)

leslie townes

how to get good data from anybody on anything related to religious belief is a challenging problem of survey design

Xander Henderson

@Jakobian I mean, we ain't scientists. :D

leslie townes

probably an unsolved one

Xander Henderson

@ModularMindset Honestly, I would expect mathematicians to be more religious, on average, than most scientists. Mathematics is not empirical, and doesn't rely directly on observation of the world to come to conclusions. Moreover, there are certain parts of mathematics that tend to invoke mysticism (e.g. lots of mathematicians seem to have a semi-mystical, Platonic view of the field).

leslie townes

19:16

if i worked in academia (particularly at a public school) and someone asked me to fill out a survey on religious belief, the first and only thing i'd wonder on my way to putting it in the trash was what pre-formed narrative they were intending to fit their survey data into. same with any other subject really, people are always trying to 'prove' that academics need more time/money, or could do more with less of those things, and most academics know this. good luck getting survey data out of them

Daniel Donnelly

Have any of you reached the mathematical plateau?

It's when you love math so much that you start to hate it again, but your learning rate actually increases because you're not over-studying, so the brain has ample time to recharge, you get a greater view of the mathematical landscape etc. 😎 The view is nice fellow maths. Don't envy me, just keep going! You will eventually get here.

@copper.hat it's all a Lie (group). I have a book somewhere called "Naive Lie Theory", got it as a b-day gift one year, I pronounced it incorrectly 😬

referring to your starred comment

pbs.twimg.com/media/E866z6RXsAE4j94?format=png&name=900x900

("mathematical plateau")

It's a thing

Lukas Heger

19:58

@leslietownes my religious belief is Grothendieckism obviously

2

Ben Steffan

I, too, believe in Grothendieck universes

acolyte of the church of large cardinals

(one may argue this is just catholicism)

leslie townes

20:19

@LukasHeger apostasy

departure from the True math

psie

21:04

If $f\in L^1(m)$, how come $F(x)=\int_{-\infty}^xf(t)\,dt$ is in $NBV$, where $NBV=\{F\in BV:F(-\infty)=0\text{ and }F\text{ is right continuous}\}$?

I think $F(-\infty)=0$ is DCT, but why is it in $BV$ and right continuous.

leslie townes

21:31

psie you are hiding the ball a bit here :) various books may define these classes slightly differently such that you can't just divorce this from its appearance in a particular book. but maybe look at folland's corollary 3.6 to theorem 3.5 for the continuity

as also e.g. suggested by folland, under the result you are asking about, referring in part for its proof to his 'proposition 3.32,' which cites his theorem 3.5

psie

ah yeah, corollary 3.6, the absolute continuity of the integral. Ok, I'll take a closer look 👍

leslie townes

yeah, the theorem 3.5/3.6 is way more fundamental than a lot of that other stuff, if i'm being honest

unless you just like knowing about versions of the FTC, which many people inexplicably do

psie

indeed. I still don't quite see from what result (theorem or proposition) we can tell that $F(x)=\int_{-\infty}^xf(t)\,dt\in BV$ whenever $f\in L^1(m)$. It feels like that should be something already proved too...hmm.

leslie townes

theorem 3.29 might help with that?

i don't mean to be rude, but why would you care

i would have stopped reading this part of folland by now :)

Sine of the Time

that's rude

2

leslie townes

21:48

some vibe things are, from radon nikodym theorem, "lebesgue integrating with respect to an L^1 function" is a prototypical example of a measure that is absolutely continuous with respect to lebesgue measure (or a "complex measure" if f is not nonnegative but just a random L^1 function)

Ben Steffan

stop reading about measure theory, read about ambidexterity in chromatic homotopy theory instead
much cooler :)

what are you measuring anyways

leslie townes

and this vibe fits into theorem 3.29 which talks about circumstances when the function "x -> measure of (-infty, x]" will wind up in NBV

ben i'm still not convinced that 'chromatic homotopy theory' has any existence separate from being a joke here

Ben Steffan

that's rude

psie

ok, thanks leslie. I will start laying the puzzle

leslie townes

well good, i meant to be rude this time :)

Sine of the Time

21:51

hope you're not colorblind if you study 'chromatic homotopy theory'

leslie townes

yeah, sounds like a very ableist field

Ben Steffan

I will split you up into your monochromatic layers and analyze your periodicity phenomena one by one

yes, and now they even demand I use both my hands :(

I will write with my left hand for the talk tomorrow to stay with the theme

what's it matter if anybody can read what I write anyways :)

leslie townes

bad luck for lefschetz

Ben Steffan

not that anybody could do that if I used my right hand either but

Transcript for

$Mathematics$ Mathematics