Mathematics - 2024-12-09

pie

05:01

:66755438, Jakobian There are a lot of thing I need to learn about then

nickbros123

08:26

@XanderHenderson I actually don't read mathematics at all. I just look at the theorem and try to prove it myself. Rarely read. Reading is tough for me, that's why my history, English, and social studies grades have been pretty bad

Also kinda why I prefer books without many words. Im not a sucker for motivation or big big paragraphs. That's why I like baby Rudin a lot

Xander Henderson

12:22

@nickbros123 I would argue that this is, often, how mathematics is read.

I will also argue that this eventually fails, because you eventually start reading work where the lines of attack are not obvious (e.g. many research papers).

Jakobian

13:51

Trying to work out all the details yourself pretty much all the time to me is a too big of a waste of time tbh

psie

14:37

I have a continuous function $f:A\subset \mathbb R^n\to\mathbb R^m$, and a cover $\{U_\alpha\}_{\alpha\in J}$ of $f(A)$. So $f(A)\subset \bigcup_{\alpha\in J}U_\alpha$ and thus $$A\subset \bigcup_{\alpha\in J}f^{-1}(U_\alpha)= \bigcup_{\alpha\in J}V_{U_\alpha}\cap A,$$by continuity. Silly maybe, but is $ \bigcup_{\alpha\in J}V_{U_\alpha}$ a cover of $A$?

I used that $f$ is continuous iff for every open $U$ there is some open set $V$ such that $f^{-1}(U)=V\cap A$.

Ben Steffan

I mean, it's probably false if the $V_{U_\alpha}$ are oranges so

psie

@BenSteffan ok, have you had a lot of oranges lately? I'm about finish by 3 kg purchase from a week ago. They're refreshing indeed (on a cold a winter day).

Ben Steffan

I have! I love oranges :)

anyways, the point is that you haven't told us what the $V_{U_\alpha}$ are so who can tell

psie

@BenSteffan ok, I will make a poster about it, take to the streets, call all my friends about it, get the message out there in the wide world...$V_{U_\alpha}$ are open sets in $\mathbb R^n$. LET'S CHANGE THE WORLD FOR A BETTER PLACE! :D

Ben Steffan

well then it's obviously false lol

clearly this is not what you want to say

Wait, let me read again

psie

14:49

Ok, maybe I should really make that poster :)

Ben Steffan

Okay, no, it's true, but it has nothing to do with openness or continuity

you're just asking whether $A \subseteq \bigcup_{i \in I} B_i \cap A$ implies the $B_i$ cover $A$, and this is clearly true

psie

Ok 👍 Where I used continuity was in writing $f^{-1}(U_{\alpha})=V_{U_\alpha}\cap A$, but ok, I think we're on the same page :)

Jakobian

15:16

@psie this depends on your definition

particularly what is a "cover of a subspace"

and if such concept was defined

Some people allow only covers of a topological space. Others, covers of a subspace

If $\bigcup_\alpha V_{U_\alpha}$ is a cover of $A$ as a subspace of $\mathbb{R}^n$

but the induced cover of $A$ as a topological space, is $\bigcup_\alpha (V_{U_\alpha}\cap A)$

hope we're on the same page here

psie

I'll have to think about it for a minute or two :)

Xander Henderson

@BenSteffan I have a love/hate relationship with oranges. I love the way they taste, but I hate the way that they leave stuff stuck between my teeth. It makes them very hard to eat (for me).

Ben Steffan

"leave but stick between [your] teeth"?

Jakobian

For example, we can define that $A\subseteq X$ is a compact subspace, if for any family of open sets $\mathcal{U}$ of $X$ such that $A\subseteq \bigcup \mathcal{U}$, there exists a finite subfamily $\mathcal{V}\subseteq \mathcal{U}$ with $A\subseteq \bigcup\mathcal{V}$.

This is definition of compactness relative to being a subspace of $X$!

But this turns out to be equivalent to $A$ being a compact space given subspace topology

psie

ah, that's good to hear

Jakobian

15:22

But those subspace (relative properties) vs space (global properties) aren't always equivalent either

We would call $\mathcal{U}$ an open cover of $A$

But there do exist people, that would rather say that $\mathcal{U}\restriction_A = \{A\cap U : U\in\mathcal{U}\}$ is a cover of $A$, and not $\mathcal{U}$

this is just difference in conventions

its a subtle difference

@Jakobian an example is, I believe, paracompactness

ModularMindset

$ 2\phi\log (\phi) K_1(2\log(\phi)) <1 $ Anyone know how to prove that analytically?

Xander Henderson

15:46

@BenSteffan stupid mobile.

They leave stuff between your teeth.

Ben Steffan

hm

I don't really have that problem

but stuff generally doesn't stick between my teeth a lot

Xander Henderson

@BenSteffan or it just doesn't bother you as much as me.

Ben Steffan

no, it would bother me

Jakobian

@BenSteffan kiwi's do that a lot

ModularMindset

16:28

Hey guys

Sine of the Time

hey

ModularMindset

I realized that I was wrong

there aren't exactly three smooth embeddings of $S^1$ onto $\mathcal I$ as I previously thought

ILikeMathematics

17:15

@BenSteffan Alright

Jakobian

18:51

I wonder if Gustin's sequence space is homogeneous

Jakobian

19:05

There should be some local properties of it, $X = Y\cup \mathbb{N}_+\times W$ that $\alpha \in Y$ has but $(n, w)\in \mathbb{N}_+\times W$ doesn't have, but I can't think of any

I've shown that every point $x\in X$ has a system of open neighbourhoods homeomorphic to $\mathbb{Q}$

this is a Hausdorff second-countable space which is not regular, for any two non-empty open sets $U, V$, $\partial U\cap \partial V$ is infinite

Oh, I think that perhaps $\partial U\subseteq \mathbb{N}_+\times W$ whenver $U$ is an open set of $X$

EM4

I have a question for integral of sin(x)/x dx from 0 to infinity

replace 1/x = integral of exp(-xt) dt from 0 to infinity.

then we use Fubini's theorem because to me it fails.

ModularMindset

19:26

hey

hi

anyone an expert in decision theory?

Sine of the Time

@EM4 So $\int_{0}^{+\infty} \frac{\sin x}x dx =\int_0^{\infty} \sin x \int_0^{\infty} e^{-tx} dt dx$

and your question is if this is equal to $\int_0^{+\infty}\int_0^{+\infty} \sin x e^{-tx} dx dt$

right?

EM4

yes.

Sine of the Time

what are the hp of Fubini?

EM4

Fubini needs to be continuous and absolute converges.

it is continuous but it doesn't converge.

Sine of the Time

19:41

why are you bounding with $1/x$?

EM4

i am trying to change the integral into a double integral.

somehow my friend's professor said it works.

Sine of the Time

ok, but why are you using $1/x$, which is not integrable?

EM4

why I am changing $1/x$ for its integration representation.

Sine of the Time

why are you bounding $|\int \dots|$ with $1/x$

EM4

user111187 (math.stackexchange.com/users/111187/user111187), Evaluating $\int_0^{\infty} \frac{\sin x}{x} dx$ with Fubini theorem., URL (version: 2015-03-04): math.stackexchange.com/q/1175145

seen this hint and see if it valid.

or maybe it doesn't work at all.

Sine of the Time

19:48

robjohn uses Fubini, take a look at his answer

EM4

is like I want to know the why behind the reason.

that doesn't explains the "why" for me.

Sine of the Time

the function $t\mapsto \int_0^{+\infty} e^{-tx}\sin x dx$ is integrable

leslie townes

EM4 as a kind of threshold matter before any specific estimates, if all you know is that |f| <= |g| and the integral of g diverges, this information doesn't tell you anything about the integral of f

at most, it tells you that you don't want to use comparison with g to assess the convergence of the integral of f

Sine of the Time

yeah that's why I was asking why did you use $1/x$ as a bound

leslie townes

yeah i am just repeating what sine has repeatedly asked about at a higher level of generality, it feels worth special mention

EM4

19:58

still confused on the understanding why it works.

even robjohn says it, I still don't see the switch of the integration.

Jakobian

I've concluded that Gustin's sequence space doesn't actually satisfy that $\partial U\subseteq X\setminus Y$ for any open set $U$

Ben Steffan

@Thorgott If you have a second, can you give me a sanity check? I'm staring at the proof of Proposition 6.10.9 in tom Dieck and can't for the life of me figure out why the special casing for $p = 0$ is necessary: You can get the required connectivity of the map $(CX \times Y, X \times Y) \to (X \star Y, X \times CY)$ directly from Proposition 6.10.1 since the maps $X \times Y \to CX \times Y$ and $X \times Y \to X \times CY$ are always at least 0-connected... right?

In fact these maps always $p$- and $q$-connected, respectively, and this gives exactly that this map is surjective on $\pi_i$ when $i \leq p + q$.

Sine of the Time

@EM4 do you know how to compute $\int_0^{\infty} e^{-tx} \sin x dx$ ?

EM4

@SineoftheTime by integration by parts.

Sine of the Time

@EM4 what do you get?

EM4

20:13

I got $frac{1}{t^2+1}

$frac{1}{t^2+1}$

my question is why is it justified for it, because robjohn also switches the integration, I don't see it.

Sine of the Time

ok, so is $\frac 1{1+t^2}$ integrable with respect to t?

EM4

in respect to x.

Sine of the Time

I think you're confusing variables of intergation

EM4

I misread your comment.

it will be arctan(t).

Sine of the Time

yes?

EM4

20:19

it will be pi/2

Sine of the Time

yes

EM4

by that we can use fubini's theorem?

Sine of the Time

you use Fubini because $t\mapsto \int_0^{\infty}e^{-tx}\sin x dx$ is intergable with respect to $t$

Ben Steffan

On a different note, are there non-contractible spaces $X, Y$ such that $X \times Y$ is contractible?

EM4

so what do I need to know for fubini's theorem then.

https://math.stackexchange.com/questions/1175042/evaluating-int-0-infty-frac-sin-xx-dx-with-fubini-theorem

so, for Fubini theorem for integration to be changed they need to be intergable then?

@robjohn I am looking at your answer just confused how it is justifed to use Fubini. Can you help this lost soul.

Sine of the Time

20:27

see here

leslie townes

16

Q: Proof that retract of contractible space is contractible

I'm reading Hatcher and I'm doing exercise 9 on page 19. Can you tell me if my answer is correct? Exercise: Show that a retract of a contractible space is contractible. Proof: Let $X$ be a contractible space, i.e. $\exists f :X \rightarrow \{ * \}$,$g: \{ * \} \rightarrow X$ continuous such that ...

ben does that answer it? aren't the factors of X x Y retracts of X x Y?

EM4

so, what do we need for Fubini then?

if integral is finite then we can switch them....?

Ben Steffan

@leslietownes ah yes of course, thanks!

ModularMindset

20:46

hey guys

wrote a quick script

check it out

think_meaning_buildß

20:56

Simple Cartoon / Animation: Man Looks Left & Right Nervously

►

Akiva Weinberger

21:24

Hey

What's new

I asked a question on main

0

Q: Taylor series has a surprising amount of powers of 10

I was messing around with Wolfram Alpha and eventually found the following surprising fact: \begin{align*}\frac{\tanh^{-1}(\frac23\sqrt{1+x})-\tanh^{-1}(\frac23)}{\sqrt{1+x}}={}&\frac35x-\frac{21}{100}x^2+\frac{303}{1000}x^3-\frac{7\,533}{40\,000}x^4\\&+\frac{437\,289}{2\,000\,000}x^5-\frac{1\,34...

combinatorics taylor-expansion hyperbolic-functions

What's up with all the zeros

Sine of the Time

"With great powers comes great responsibility"- uncle Ben Steffan

Ben Steffan

nevermind, classical case of "old man doesn't get pop culture reference"

ModularMindset

This Cylinder Will Change How You See Geometry

►

Claudio

is this a standard result I don't remember: if $M$ is a real positive-definite symmetric matrix then $M = R^T R$, for some non-singular matrix $R$?

Sine of the Time

21:43

@Claudio this?

Claudio

@SineoftheTime I'm reading this,

but it doesn't say anything about the nature of $R$ apart from it being non-singular

but I guess it might be yeah

leslie townes

claudio: there are a lot of results that imply that kind of thing, but none that i know if with a canonical name

Sine of the Time

@Claudio I see

leslie townes

and for the sake of the cosmos please do not personally write anything that could be read as implying that there is canonical name for that result, navigating the world of matrix factorizations is confusing enough as it is

Ben Steffan

ah, if it isn't the famous $RTR$-decomposition

Claudio

21:49

I won't go deeper Leslie, I won't

leslie townes

if you impose the triangularity structure, people would probably associate that with the cholesky name, but if it isn't important to your application then i wouldn't use that name

Sine of the Time

This is the Leslie-decomposition

leslie townes

and i'm certain you can find books phrasing the hypotheses or conclusion of "the cholesky decomposition" differently, to a degree that i would probably avoid using that terminology in any situation where details might matter

Claudio

yeah yeah I understand, I've seen matrix theory books

and I didn't like what I saw

@leslietownes thank god I study physics :)

leslie townes

there's a lot of stuff like this, where people will say "the unitary from the singular value decomposition," well which of at least two do you have in mind, and surely you're aware that "the" unitary you're thinking of might be the adjoint of "the" unitary i'm thinking of, and, just, why do we do this to each other

but more seriously if you have something that could make use of any factorization of the form R^T R and you call it by a name associated with a more specific factorization, peolpe who read the result might be confused and think that you "need" or are at least using the extra structure that they associate with the name

Claudio

21:53

I can feel your pain through the screen

leslie townes

which can cause people to misapprehend your results and maybe even misattribute them later

Claudio

I'll follow your piece of advice

leslie townes

i'll take my pills and calm down now :D

Claudio

lol

Jakobian

@leslietownes I had you figured for an algebraic topologist...

I still haven't figured if Gustin's sequence space is homogeneous or not (most likely not). Its hard to come up with anything

Thorgott

22:34

@BenSteffan Prop. 6.10.1 is only stated for $p,q>0$ (because he phrases it slightly awkwardly in terms of relative homotopy groups), but I believe that the statement still holds for $p,q\ge0$ and then immediately applies to yield Prop. 6.10.9 as you say

Ben Steffan

ah, yeah

I found that out in the meantime as well, forgot to let you know

Prop. 6.10.1 should state what the assumptions on $p, q$ are

luckily I have the stronger version in the lecture notes I'm working with, so I don't need to special case anything

although one can have fun with the special case :)

Thorgott

@BenSteffan it does by saying $0<i<p$, no?

Ben Steffan

that constrains $i$, not $p$

it could allow $p = 0$, then the condition would be empty

Thorgott

I suppose that's fair

the issue is that it's the wrong condition for $p=0$

Ben Steffan

...some people do it like that

yeah, I know

I'm kind of surprised that it doesn't demand $\pi_0(X, A) = 0$

but I guess everything works out without that

Thorgott

22:42

I think you can just restrict to the components containing the basepoint and then everything works out

was just explaining earlier today to some students why SvK does not require $U,V$ to be path-connected for the same reason lol

Ben Steffan

right

Thorgott

22:57

actually, I'm not sure what the right conclusion for $p=q=0$ is

I don't think I've previously come to fully appreciate the difference between assuming connectivity and vanishing of relative homotopy groups in the hypotheses of Blakers-Massey

Ben Steffan

I commend you to Proposition 11.9 in him-lueck.uni-bonn.de/data/script_AlgTOP_I_www.pdf

Thorgott

yeah, sounds about right

I'm not used to relative $\pi_0$ somehow

Ben Steffan

I guess it's a matter of taste

Thorgott

yeah, I think the script you linked is a slightly cleaner exposition of the version in tom Dieck

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