12:00 AM
Huh, how did a trained medical professional end up hurting you? :)

there are some brusque movements to get things in line ... very different style from the guy I went to for 15 years in GA.

Have you tried seeing a physical therapist instead?
Im 6' 5'' and have had terrible posture my entire life and quite a lot of pain in my back and chest over the last year, but it has improved remarkably with exercise and massage.
in PT for about 5 months.

I never have tried PT, because I had such fabulous fortune with the guy in GA. We'll see how this develops.

Well the physical pain at least

Back and neck pain are the worst.
I wonder how @AlexGruber is doing. Haven't seen him here in ages.

12:10 AM
There was a point when I was trying to get an appointment to see a doctor on my university's health website about my muscular chest pain, and couldn't do it because on the forms to make an appointment chest pain would immediately direct to a dead end page telling you to go the ER.
So i think I just lied about my symptoms..

uh huh, cuz of heart.

Well if its been persistent for 4 weeks with normal vitals and no history of heart problems, youre going to be waiting a while in the ER.

I've had two major heart surgeries, too, so I am acquainted with the origins of the word "heartburn."

Hi @TedShifrin

hi @Karim

12:17 AM
how is teaching student @PVAL
I have never done it before

@Karim Often soul-sucking. Always exhausting. Usually fun.

Wow, I don't think in all my 40+ years of teaching I would have used some of those.

@J.M.isback. that doesn't sound like fun

Indeed not, @robjohn

@TedShifrin Often my job is to tell people formulas they have to memorize without giving any good reason to other than the scholastic one(usually I don't have these formulas memorized). It's hard to be terribly optimistic about this part of it.

12:26 AM
Yeah, @PVAL, this is one on which I'm slightly conflicted. But shouldn't a college math student know basics like area of a circle and volume of a box ... and have some idea about units?

$(0,\pi]$ is open in $mathbb{C}$ right?

What?!

@TedShifrin Perhaps, but memorizing something like the general solution to the Euler's equation when the characteristic equation has a repeated root seems kind of meaningless if you don't know how Euler derived it.

Yeah, actually, I think most DE courses don't give any motivation for what happens with repeated roots of the characteristic equation, in general. That bugs me.
(No, I don't want to do matrix exponentials.)

?
to my question ?

12:30 AM
@Karim: You should spend more time on understanding $\Bbb R$ and $\Bbb R^2$ than on all this formal set crap.

which set @TedShifrin ?

all the stuff you were talking about last night

yh

Yo guys

hi Lucas

12:33 AM
Wassup?

You realize, @Karim, that $(0,1]$ isn't even open in $\Bbb R$?

yeah

And no interval in $\Bbb R$ can be open in $\Bbb R^2$ ?

yes
yes sorry for some reason I my mind fucked up for a sec

ok, then

12:35 AM
that is because we can never find a open ball around that set that contains

NO ... say it right

any point in (0,1]
that is it there doesn't exists a real number $\epsilon$ such that given any point x in (0,1] d(x,y) < $\epsilon$
the metric we use in $R^2$ is the euclidean one

NO, @Karim, you need to concentrate on basic stuff. Get it right.

that if every point has a neighborhood lying in the set.

@TedShifrin Or the motivation for $u_1'y_1+u_2'y_2=0$ in variation of parameters. I don't even really understand the motivation for that...

12:40 AM
Oh, @PVAL, I used to know that. I don't think that's so hard to motivate.

I have a question about sets. I think it's pretty hard for highschoolers. I was given a physics exercise that states $5t^2 -20t + 2d > 0$ where $0 \le t \le 4$ and $d > 0$. I must find the minimum value for $d$ given the conditions above

Graph the parabola, @Lucas.

But I don't have d :(

What's the meaning of $2d$?
I don't mean on a computer, @Lucas. I mean using your brain.

d is the space between two cars. The original equation was $S_A < S_B$
Oh, a constant
That way you state, using graphs

12:42 AM
If we're talking about graphing $y=5t^2-20t+2d$, where in the graph do we see $2d$?

y axis, x = 0

Right.
Now figure out where the vertex of the parabola goes.

but then, if we get t = 0, d > 0
which is obviously not the answer

Yes, as I said: Figure out where the vertex of the parabola is.
@Karim: It is essential in analysis/topology to get the quantifiers correct and the sentences carefully written.

yeah

12:46 AM
So I get its derivative and say its zero?

OK, you can do that, or you can use algebra and complete the square. Either way.

Just to state: We got a car at 20m/s and another car at 10m/s with a distance of d. If to not crash, car 1 have an acceleration of -5m/s², what is the minimum value of d?
(to not crash)

I'm not paying attention to that. You need to solve the basic problem about the graph of that quadratic.

Then 20t - 2,5t² < d + 10t

Huh?

12:50 AM
It's alright, I'm just showing what I thought
These are the spaces. Integral of velocity.

Where should the vertex go if you want the smallest $d$ so that $5t^2-20t+2d\ge 0$ for all $0\le t\le 4$?

It's only greater (cannot be equal), otherwise the cars collide

Well, then there is no smallest $d$.
Typical sloppy physics talk.

Hi everyone. Does anyone have a minute to talk about an old real analysis question on the site?

Oh, what the hell. I'm confused

12:52 AM
They need to ask, what's the smallest $d_0$ so that whenever $d>d_0$, ...

lol

@morphic: Don't laugh at me. I'm not going to talk about inf.

Thanks for the time man. I'll take a look at the answer and try to understand

Answer my question first, @Lucas. And then think.

@RudytheReindeer Hi, I see you around the site a lot because for some reason whenever I search for questions, you have always asked the same ones I have now

12:54 AM
Isn't it a few months early for Rudolf the Reindeer? :D

@TedShifrin Where I live it's always Christmas : )

You have internet at the North Pole?
So what's the question, Rudy?

@TedShifrin Sure we do.
@TedShifrin This one.
I think neither of the answers actually answers the question. But on closer inspection the question doesn't make sense. At least not to me.
After all: what does it mean for the x limit to exist? It seems to me that the x limit is a function.

The answer seems to be doing the converse, not the original.

So perhaps the OP meant to say that the x limit is a function?
Exactly!

12:57 AM
I think the question was: If the limit exists, then show that you can get the same limit first letting $y \to b$ and then letting $x\to a$, and oppositely.

I see. So the claim if the joint limit exists then so do the iterated limits is true?

Robin Chapman's counterexample is good.
No, it's false.

Sorry, I misinterpreted your previous comment.

I was restating the original question, not saying it's correct as stated :)

I wanted to edit the question to make sense.
Ah.
But I thought the question was trying to ask the following:
If the joint limit exists and each single variable limit defines a function then the iterated limits exist and are equal.

1:01 AM
@ted they said that another condition not to collide is that $v_A = v_B$ when they are in the same space. Didn't get it.

Do you know if this would be true?

Oh, good point, @Rudy. In Robin's counterexample, those individual limits don't exist.

Btw the vertex is when t = 8

$t=8$, really?

Oh gosh. t = 4

1:02 AM
Nope.

OH GOSH

LOL

t = 2

OK, better. :)
Now what must be true about the vertex of the parabola?

That's what happen when you forget the power rule :P
Let me guess
Without 2d we get -20
Then d > 10

1:04 AM
OK.

Marry me, sir.

Sure, any time :)

Thermodynamics is so boring... :/

@Rudy: Seems correct if one assumes those individual limits exist. Just write out a proof.
I like thermodynamics.

You're not my fiancé.
:p

1:07 AM
Whose fault is that?

?
Sorry, didn't get it

Don't worry :)

( give me a break, I don't even speak English :P)

Apparently not. I don't speak Portuguese.

Thermo is like : What is the final composition of the system
And I'm like: "How the hell am I supposed to know"

1:09 AM
Or what temperature is the most comfortable

Oh, you know I'm Brazilian then

I looked at your profile page, and it was full of Portuguese.

Oh yeah. You got it.

Somehow in topology class we got into a discussion of whether temperature is continuous or not

one posits that it is, @morphic.

1:11 AM
Man, calculus is the best thing on mechanics
You only need to remember 2 formulas
(formulae, IDK)

well, but it does help to remember calculus correctly

@TedShifrin But those limits are functions. Saying they exist makes no sense, right?

I gotta sleep

night, @Lucas

1:14 AM
Thank you all. Night!

Sure it makes sense, @Rudy.

@TedShifrin Better question: if $f(x,y)$ is a function, does it imply that $f(0,y)=g(y)$ and $f(x,0) = h(x)$ are functions, too?

We're saying that for each fixed $x$, $\lim\limits_{y\to b}f(x,y)$ exists.

Oh.
Ok.

Not a better question, @Rudy. Of course it does.

1:19 AM
Then I think the OP wanted to ask that. If the joint limit exists and for all fixed x the y-limit exists and for all y the x-limit exists then so do the iterated limits and are equal.

I think that is what he asked in poor English.

But then neither of the answers answers the question. Maybe I'm missing something.

You're correct.

I am going to edit the question.

I haven't looked super carefully, but I think you're right.

1:20 AM

I still say you should figure out a proof :)

OK. I agree. Maybe I should do that and then post it as an answer to the edited question.

I would encourage you to do that.

Do you agree with the edit?

NO, @Rudy. Those limits need to be functions of the remaining variable.

1:25 AM
@TedShifrin The other variable is fixed. So the result should be a constant.
I'm confused.

But $L'$ and $L''$ depend on $y$ and $x$, respectively.

@TedShifrin Yes, what's wrong with that?

The way you've written it, it seems that they are universal constants.
Just say that for each $x$, $\lim_{y\to b} f(x,y)$ exists , etc.

Ok, I'll add an index to each.

I'm outta here. G'night for now.

1:28 AM
Good night!

196884 = 196883+1

I am debating whether I should be the Monster for Halloween.

Conway says his Halloween parties are fun

are harmonic
nvm

I have a question about partial ordered chains and vectorspaces
It is in the proof of Hanh-Banach theorem in a textbook
They define $g\leq h$ to mean $h$ is an extension of $g$, i.e. $D(h)\supset D(g)$ and $h(x)=g(x),\forall x\in D(g)$
Then they deduce that any chain is a vector space

1:44 AM
"any chain is a vector space" - a chain is a chain. can you quote the conclusion verbatim?

the union of the domains of the functions in the chain is a vector space
that's hardly the same as "any chain is a vector space"

Sorry, I did plan on copy and pasting
That's the two lines I cut out
So the union of the domains should just be the domain of the top domain in the chain right?

if the chain has a top at all

Okay, so the point is, we haven't used zorns lemma yet, okay, but how is it a vectorspace

1:47 AM
you can add any two things in it and scalar-multiply anything in it

Why?

say you have two things in it. figure out why you can add them.

Because they are linear functionals, thanks

no...

no?

1:54 AM
If I have vector spaces $V_0\subset V_1\subset V_2\subset V_3\subset\cdots$ then can you see that $\bigcup V_i$ is a vector space? It has nothing to do with what types of things the vectors are (besides the fact they are elements of vector spaces that are in a chain like that)

Yes I can
So each of my domains are vectorspaces was what I didn't realise

2:09 AM
Yes thanks after that the rest of the proof was easy to understand

4 hours later…
5:51 AM
Hello, anyone up?

@sanic i don't think so

me
hi @SanicHodgeheg

Hey @Ramanewbie and @Huy. I was just wondering whether it's valid to substitute the set $A$ with the logic formula $x \in A$ and work with an expression of sets as a logic formula instead.

wat
logic
?
can you give an example

$A \cup B \cup C = (x \in A) \vee (x \in B) \vee (x \in C)$

6:01 AM
@SanicHodgeheg No, that is not correct
you need to have an $x\in$ on the left too

you should have a $x \in$ yea
what he said

@TobiasKildetoft Makes sense. Once I have that defined though, I can use the right hand side directly.
Basically, am I allowed to translate a set theory problem into a logic one in that way?

@Tobias: Are you doing a PhD or what are you doing atm?

@Huy I am a postdoc

ic. can you somewhat summarize what you're working on for someone not in your field? @Tobias
doesn't have to be perfectly accurate ofc

6:06 AM
@Huy I actually work on a couple of different things. Continuing on from my dissertation, I am working on figuring out how certain tensor products of representations for algebraic groups decompose

ic

I am also working on understanding the 2-category of Soergel bimodules in type $B$, hoping to understand if all simple transitive representations come from left cells
and parallel to that I am working on understanding how a new description of special representations fits into various settings

unfortunately I hardly know anything about representations and bimodules :(

@Rememberme Hi

6:08 AM
too much algebra for me ._

@Tobias Mind checking a proof of mine?

@Huy Well, the category can also be described as the category of projective functors on a certain other category
@Rememberme But the things asked about being even in the question is not itself a permutation

^
(is it common to write $Z_2$?)

@Huy It varies a lot between areas

yeah, I've never seen that before

6:11 AM
Is there any problem writing $Z_2$?

I remember having an algebra lecturer who insisted on using it even though it is going out of style, since he had been part of introducing the notation

not if you explicitly state what you mean by it, I was just wondering if that's the usual notation because I've never seen it before

@Rememberme Well, the same symbol is also used for the $2$-adic integers
and there is no other notation for the $2$-adic integers

So should I write $\{-1,+1\}$ then ? I guess that will solve the problems

@Rememberme But it will not really solve the problem of the disconnect between what you calculate and what the question asks about

6:13 AM
@Tobias: Wouldn't you write $\mathbb{Q}_p$ for p-adics?

@Huy Those would be the $p$-adic numbers, with ring of integers $\mathbb{Z}_p$

hm, I need to quickly look them up, else I just keep confusing things

the $p$-adic numbers are the Cauchy completion of the rationals with respect the the $p$-adic metric

yea

the $p$-adic integers can be described as a certain limit, and has the $p$-adic numbers as its field of fractions

6:15 AM
aha
I don't think I've seen p-adic integers in my study actually
only the numbers
wait
is $\mathbb{Z}_p := \{x \in \mathbb{Q}_p|\, |x|_p \leq 1\}$ p-adic integers or is that something completely else?

@Huy Those are indeed the same (I don't recall how easy that is to show)

@Tobias What is a functor?

@Rememberme a morphism between categories

I am asking this question because according to Balarka there is a easier proof that a topological group is abelian using that $\pi_{1}$ is a functor@Tobias

I think we only used that set (and not the notion of integers) to prove that for any p-adic number exists a unique sequence $(a_n) \subset \{0, \dots, p-1\}^{\mathbb{N}}$ with $$x = \lim_{n \to \infty} |x|_p^{-1} \sum_{k=0}^n a_k p^k$$

6:20 AM
@Rememberme But a topological group need not be abelian

I mean a topological group with loops having a base point at the identity element of the group@Tobias

@Rememberme I have no idea what you mean by that

which topological group doesn't have some loop with base point at $e$?

Oh my bad . I am sorry
I meant that the fundamental group of a topological group is abelian@Tobias
I mean to say that if $G$ is a topological group with $x_0$ as its identity then $\pi_{1}(G,x_0)$ is abelian where loops have $x_0$ as its base point@Huy

@Rememberme But the torus is a topological group

6:28 AM
math.stackexchange.com/questions/686496/… This is the question I am trying @Tobias

@Rememberme Ahh, I see
But isn't the torus a topological group?

@TobiasKildetoft: Why do you think its fundamental group isn't abelian?

@Huy Because it is free of rank $2$ (or am I just misremembering this?)

the fundamental group of $S^1$ is $\mathbb{Z}$, so the one of the torus is $\mathbb{Z} \times \mathbb{Z}$

@Huy Ahh, so I was being silly
@Rememberme anyway, the proof that uses functors also uses some pretty deep ideas from category theory

6:34 AM
I think you can also argue with the universal cover very quickly that the fundamental group must be abelian, but my knowledge of covers isn't very good so I'd have to write it down on a paper. :P
@TobiasKildetoft: do you by any chance know whether there's some similar way to easily figure out the fundamental group of a quotient, similar to the product?

@Huy I doubt it, as it will depends on how the subset is "positioned", not just on what the subset looks like
unlike the case for products

yeah, that's what I thought
would be quite useful though :P

though there are probably spaces which are "uniform" enough that the positions does not matter
but I would not want to try to make that precise

ok. I'm off for some grocery shopping and then I'll finally start revising functional analysis

7:43 AM
@Huy MY approach has been to take two loops , use concatenation of loops and then prove that it is abelian. I would like to see the universal cover method

7:53 AM
imgur.com/gallery/2IMgS I thought these were real (and Queensland Rail was another British train franchise) until half way down.