erm..

Comment va?

Right?

Comment ça va @Studentmath

:(

@Studentmath That being said, some people in the South say Comment va

Fromage.

The southies are considered better or worse?

Omelette de fromage !

@Studentmath WORSE (I'm from Paris :P)

:(

Paris's French is real French

what just happened?

I see a ton of trash talk for transcripts and no math anywhere.

@Alizter smbc-comics.com/index.php?db=comics&id=2245#comic

@Balarka I can't prove what I wan't to :(

I can prove something really close

WAT

In a very simple way compared to the long, terrible proofs of the usual case

I am too busy with topology

Oh you can't really help me with this

I just wanted to whine to someone

LOL

Where are you in topology?

closed sets.

Hello @ccorn

@BalarkaSen Hi

OEIS debugging

It looks like @Anastasiya-Romanova didn't get my point here ...

I mean she corrected her work, but she didn't continue ...

What's the natural way to go there?

dilogarithms, differential equations, auchhhh, I begin to have headache ...

Just post the solution you have in mind @Chris'ssis. As an answer.

I need to parameterise the curves of intersection of $x^2+y^2+z^2=4$ and $x^2+y^2-z^2=-2$. I've got that one of them is $r(t)=(cost, sint) \forall t \in [0, 2\pi]$, but how do I find the other one?

@BalarkaSen What would you do?

I wouldn't even try to help :P

Even if I knew the answer.

@BalarkaSen Are you upset on her?

@user112495 (cos,sin) is in R^2 not R^3. think of those two equations as a linear system in two variables and solve for those two variables; what subexpressions am I referring to as the variables?

No. What made you think that, @Chris'ssis?

I mean I will try to spend my time studying something worthwhile than helping people around.

@BalarkaSen Just asking.

@BalarkaSen Ah, selfish ... :D

No, reasonable.

@BalarkaSen Just kidding ...

I can't possibly help people with half knowledge.

;)

@anon!

I am topologizing!

@anon Ah, so would it be $(cost, sint, \sqrt{3})$?

@user112495 yes, that would be one of the curves

@user112495 Answer anon's question. It's exactly to the point.

what other value for z do you get besides +sqrt(3)?

@anon $(cost, sint, -\sqrt{3})$

right

@BalarkaSen topologizing what?

Oh, that's not as bad as I thought it was going to be haha. Thanks!

@anon I am studying topology.

Finally.

I wants to know a nice system of representatives for the orbits of hyperbolic triangles under the action of PSL2(R)

ah

@anon Hyperbolic triangles as in $\mathcal{H}/PSL(2, \Bbb Z)$?

@anon $\mathbb{R}$? Not $\mathbb{Z}$? Alas. I have written an asymptote script to generate such tilings.

Oh noes R

@ccorn yes R

@BalarkaSen as in $\cal H$

@anon The usual (2, 3, infinity) triangles?

IOW a system of representatives for (triangles in $\cal H$)$/{\rm PSL}_2(\Bbb R)$

@BalarkaSen what do you mean by the usual triangles?

@anon If possible would you be able to help me with this other unrelated question. I need to prove that $\lim_{t \to \infty} \int_a^b sin(tx) dx = 0$ We are currently studying step functions and regulated functions, and so I believe I have to involve these in my answer.

Do you mean the triangles as in the ones appearing in the hyperbolic geometry on H?

three points in H connected by three geodesics

ah

I am using the upper half plane model I guess I should add

right

@user112495 you're not allowed to just compute the integral?

you should know the antiderivative of sine

Hello @Kaj

@anon I don't think so. Ah, so shall I just use the FTC?

Hey there Balarka

@user112495 sure it's easy with FTC

@Pedro!

Hello.

How's the studying?

Very good. I am doing closed sets.

Ew. I don't find cosed sets attractive.

$\cos(\Bbb R)$ is boring, yes.

@PedroTamaroff Have gone through the proof of Hausdorffness of metric spaces.

Not a big deal, it was set up as an exercise in Simmons.

Yeah, it is pretty trivial stuff Bal.

It is not bad to think about Hausdorffness as a replacement for the triangle inequality.

Oh?

Yeah. It is merely a piece of intuition. But lots of math is having the right intuition built up.

I'd have to go now.

Byes.

I have an exercise for you. Prove that the metrics $d(x,y)$ and $\min(d(x,y),1)$ generate the same topology. That is, a set is open in one iff it is open in the other.

Ta ta.

@PedroTamaroff OK.

@anon So, given two triples of vertices in $\mathbb{H}\cup\{\infty\}$, you want to know whether these generate the same $\operatorname{SL}(2,\mathbb{R})$ orbits?

yes

Well $\bar{d} \leq d$, @Pedro, so trivially there is a $d$-ball contained in a given $\bar{d}$-ball. We have to prove that there is a $\bar{d}$ ball contained in a given $d$ ball. $\{x : d(x, y) < r\}$ be our $d$-ball. WLOG assume $r < 1$. Then $\bar{d}(x, y) < d(x, y) < r$ so $\{x : \min(d(x, y), 1) < r\}$ is our $\bar{d}$ ball. right?

Oh wow in fact balls in $(X, d)$ and $(X, \bar{d})$ coincides.

@anon What about fixing a fourth point $D=\mathrm{i}\infty$ and using the three hyperbolic area measures (using the $1/y^2$ metric) of $ABD$, $ACD$, and $BCD$? The areas are preserved under $\operatorname{SL}(2,\mathbb{R})$, and if I have counted right, three real parameters is precisely the number of degrees of freedom that should remain.

remain after what?

@anon to identify the particulars of the orbit of a given triangle.

are you saying the orbit of ABC is determined by the multiset of area measures {ABD,ACD,BCD}?

@anon Not sure yet. But conversely, the area measures are determined by an arbitrary triangle from the orbit. Distinct area multisets imply distinct orbits.

right

Hi @anon @Hippa @Balarka

Hello @TedShifrin

I am noticing that my position as an insignificant rat has improved among the analsysts and geometers in this chat.

Possibly because of my decision of studying topology instead of calling it BS?

@anon scrap that, $D$ would have to be transformed along with $ABC$. But it might be something along those lines.

ah right

Insignificant rat? Nah, just a self-centered child :)

which is better than the adult version

Especially Pedro saying hello to me was new.

LOL ... @skull: Was that remark aimed at anyone in particular?

seemingly :P

nah, just in general };-)

Thanks for sharing, @skull.

np pal

Pal? Right.

LOL

:D

@Balarka! @Ted! @Ice!

Heya

Hi!

Int(Q) is null? Why? blank stare

Q ain't contain no R-balls

I need to sleep anyway. byes.

@anon oh?

wait let me think about that

I shaved for the first time today

hmph grmph yeah. you can always produce the irrationals, is that right?

those which are in the R-balls but not in Q

@anon

hi @Studentmath

How's it going?

One of my students asked a cool question. We had an exponential random variable $X$ representing the time before a computer crashes, with mean $c$. He asked how many expected crashes you would have in time $c$. :)

:)

Hi @Ted, I am Mr. Eyeglasses

LOL ... hi mr Eyeglasses/nabla ... aren't you old to be shaving the first time? :)

@TedShifrin time being the cardinality of the continuum?! :O

Assuming you mean $\Bbb Q\subset\Bbb R$, @Balarka, it has empty interior, yes.

@TedShifrin I am trying to visualize why

My face is bleeding :(

If your universe is $\Bbb Q$ (with the usual topology), the answer is different.

no it is R

of course interiors differ as the universe varies

Well, no open interval in $\Bbb R$ around $0$ contains only rational numbers. You of all people should know that :P

@user130018 splash your face with cold water

I put Vaseline on it

@TedShifrin in my defence, i am sleepy

Go sleep, @Balarka.

it's 5:18 of the midnight here

I'll go to sleep when the hell freezes over, @TedShifrin

:P

It is 7.49 am here, and I was awake the whole night, lol.

that's just nothing @Jasper

I wonder if @Studentmath is pondering my question ...

I have stayed awake for two whole days

@Ted I actually am

cool :)

I did figure it out, btw

@BalarkaSen Are you on drugs?

But I thought it was a good question.

no @Jasper

It's a great question

it was a crazy sleep disorder back then

partially because of staying up late at night too much

Hi @JasperLoy

@user130018 Hi, do I know you?

hi @Jasper

@JasperLoy Yes

@user130018 Hi Bart, LOL.

lol

@Ted well you can always go back to the definition, and fine the expectation by integrating between 0 and $c$

but we want expected number of crashes .. this is a new random variable, @Studentmath. We need to use a model that when the computer crashes, we reboot and get the identical random variable $X$ for time to next crash.

@TedShifrin Damn probabilists are sick.

heya @Pedro :)

@pedro What is your BMI?

I think @skull called me a self-centered old guy above ...

@Pedro did you see my solution?

@TedShifrin Hello Ted. How was dinner?

@TedShifrin you? nah.

'twas fine, @Jasper, thanks ...

My BMI now is slightly over 25, lol. Need to lose some weight.

@JasperLoy Oh. That.

I have no idea.

@Pedro is too thin :P

LOL what were you thinking about, @Pedro?

But honestly this BMI thing is just a very rough guide.

@TedShifrin WHAT.

It doesn't mean too much.

@Pedro, why are probabilists sick?

@Ted, hmm, right. What's the answer? (not the solution way, just the numerical solution)

@TedShifrin They always want to calculate probabilities of catching deadly illnesses or getting killed in a crash?

@TedShifrin I would never even think of doing such a thing Professor.

What's your intuitive guess, @Studentmath?

That's because insurance company hire actuaries and pay them zillions of dollars, @Pedro.

For some reason, $c$

@TedShifrin Currently I'm weighing 73/74 kg.

Wait, erm

@PedroTamaroff What is your height?

I am 30.

Blah.

@JasperLoy 1,80 m, give or take.

I'm about 75, @Pedro, but you're a foot taller.

@Jasper He is an Argentine. Of course he is not tall.

No, not a foot ...

@BalarkaSen WAT,.

You certainly didn't mean that, @Studentmath.

Wait you ARE tall, @Pedro?

BMI=22.7 then.

@BalarkaSen Do you even metric system Balarka.

@BalarkaSen @pedro is tall, dark and handsome.

I do metrics, no metric system

@Balarka smirked when he said that ... and then he fell asleep.

@JasperLoy I'm not dark. What the hell.

With mean $c$ that means we expect one crash every $c$ time, so I would expect $1$ crash by time $c$

@BalarkaSen 6 feet.