Mathematics

Not quite, Cookie. I'm not sure what you mean.

I have no idea what Semiclassic is talking about.

@Semiclassical Who?

oh, right

I missed whatever @Cookie just typed.

5:06 AM

@akiva I didn't realize you were responding to his message on chat. I thought you were responding to this

@TedShifrin needlessly convoluted counter-trolling for my own amusement.

@TedShifrin I don't want to continue past derivatives because integrals are very ugly

@Semiclassical I did not see that

I love integrals, Meow.

yeah, without that this conversation doesn't make a lot of sense

derivatives are like smashing a glass table and then integrating is putting it back together with gum

5:08 AM

ehhh. numerical integration is like that

That clears a lot up for me, I was quite confused

@Ted I'm not sure how I'd put it technically, but $s(\textbf{v}-\textbf{u})$ and $t(\textbf{w}-\textbf{u})$ are both lines, and $s(\textbf{v}-\textbf{u}) + t(\textbf{w}-\textbf{u})$ then forms a "plane." So then the vector $\textbf{x}$ is that "plane" situated at the tip of $\textbf{u}$

Oh, interesting. Semiclassic, I tried to do the 3D version by integration and failed. Affine geometry for the win.

symbolic integration is more like putting a puzzle together

That is very assish

5:09 AM

If I'm not making any sense, thats totally fine

yeah

this is also why I asked about plagarism policy earlier :3

So it's the plane through the three points, Cookie!

Symbolically, differentiation is easy and integration is hard. Numerically, integration is easy and differentiation is hard.

^

Most functions have no antiderivative

in elementary terms

Numerical diff is fine, DogAteMy.

5:11 AM

ehh. compared with numerical integration, numerical diff is harder

@Ted I know its not as important as the concept, but is there a "nicer" way to refer to a "line" $a(\textbf{v}-\textbf{u})$, for $a$ $\in$ $\mathbb{R}$ and $\textbf{u},\textbf{v}$ $\in$ $\mathbb{R}^{3}$?

@TedShifrin Is it?

the main exception to 'symbolic differentiation is easy' is how tedious it can be. the other one is that transcendental functions can have derivatives which are also transcendental functions

You just do discrete differences ...

I heard something about $\frac{f(x+ih)-f(x)}{ih}$ being surprisingly stable (for functions that can be analytically continued to the complex plane), actually

5:13 AM

why arent both easy numerically?

Line “spanned by” ... Cookie. You'll get to that in a few sections.

but just doing $\frac{f(x+h)-f(x)}h$ or $\frac{f(x+h)-f(x-h)}{2h}$ doesn't give you good enough approximations, or so I've heard

I dunno, I guess I'm probably wrong

main issue I know of is oscillations

Gotcha, I'll keep any eye out for it. Thank you

Like it doesn't converge fast enough

5:13 AM

why arent both easy numerically?

wow nice internet

can you not just add $f(t) dt$ for integrals, and calculate average rate of change for derivatives?

some discussion here: mathoverflow.net/questions/64302/…

People do graph-theoretic Laplacians, for example.

@MeowMix In a word, stability

main thing I'll note there is that both of the specialized techniques they offer amount to turning differentiation into integration

why do mean values have their own theorem if theyre so mean

5:17 AM

Past Meow's bedtime!

im just wondering

amusingly, there's a concept of 'amenable groups'

and roughly speaking that corresponds to 'groups for which it makes sense to talk about mean values'

so having mean values makes you amenable.

You get a lot of colorful terminology from recreational math as well (see amicable numbers)

i always find that amusing.

(and now it's past my bedtime)

Incidentally, that's how I first learned what the word "amicable" means

5:18 AM

Happy number

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits in base-ten, and repeat the process until the number either equals 1 (where it will stay), or it loops endlessly in a cycle that does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers (or sad numbers). == Overview == More formally, given a number n = n 0 ...

@Ted I need some cool limits

one cool one was $\lim_{n \to \infty} (n! / n^n)^{1/n}$

$$\lim_{x\to 1}\frac{\zeta (x + \sin (x))}{x^2 \zeta(x)}$$

(I am rambling)

What's the double limit that gives you $1_{\Bbb Q}$

Incidentally, what's the name of functions of the form $1_A$? I'm forgetting

dirichlet?

indicator?

5:21 AM

That one, thank you

which, lol

@MeowMix What's this? I wonder if it involves $\pi$ in the end

oh hey, that looks a stirling problem

Hence my suspicion of $\pi$ happening

'cause $\pi$ shows up in Sterling

nah, I expect $e^{-1}$

5:23 AM

¿Por qué no los dos?

@TedShifrin hi

Because the factor of pi in Stirling is a constant

@Semiclassical Oh turns out you're right

@Semiclassical Mm, so it gets killed by the $1/n$ exponent

I gave a geometric ine in here a whike ago, Meow, which DogAteMy did.

right

it's in the pre-exponent, so it's irrelevant

that's a phrase you see a lot in semiclassical stuff, fyi

estimates of tunnel splittings usually look like $\Delta \sim e^{-S}$

5:26 AM

Akiva: Previously I was trying to visualise the tychonoff plank, but then I realised I am so bad at reasoning with closed sets that I cannot even verify my diagram, thus I postponed the stuff later

and one usually doesn't bother to worry about the preexponential factor because $S$ is so much more influential

@TedShifrin A geometric what?

(And how long is a whike)

@TedShifrin

Limit problem ... ratio of areas of two triangles as angle squeezes to 0.

Karim ... no need to keep pinging. It's obnoxious.

okay sorry

Jeff

5:31 AM

is it true that subspaces P and S are isomporphic if both have a trivial intersection with another subspace Q? It feels like this shouldn't be true if you consider R^3 and have Q be a one dimensional subspace P a two dimensional orthogonal subspace and S a one dimensional orthogonal subspace.

Right, Jeff.

In higher dinensions, even worse ...

Right that it's not true, or right that it's true?

Morning everyone!

His counter is right.

Quick question, is there anything special about an open set of a compact space as opposed to an open set of an arbitrary topological space?

Jeff

5:36 AM

It's contained in a compact set? Just guessing.

It's closure would be compact, but apart from that I can't think of anything

@Perturbative Intrinsically, nah, 'cause everything's contained in its one-point compactification anyway

(and is open in it)

@AkivaWeinberger Could you elaborate a little further please?

Because there's this problem that I'm working on which says "$X$ is a locally compact, Haursdoff space if and only if it is homeomorphic to an open set of a compact Haursdoff space"

OK, so we know it's Hausdorff

Yeah that part is trivial

5:44 AM

Are there problems with taking one-point compactifications of non-locally compact things?

It wouldn't surprise me, but it would also surprise me

What's an example of something that's not locally compact, actually?

Oh, $\Bbb Q$

@AkivaWeinberger The only place I saw that deals with non-locally compact things is nlab

ncatlab.org/nlab/show/one-point+compactification

Also Wikipedia lists the origin union the open upper half plane

but $\Bbb Q$ is particularly simple

@Akiva Maybe this wouldn't give you something that's actually compact

?

Out of the bushes, a wild Daminark appears

If we're talking about metric spaces which aren't locally compact (and I think are thus not complete), adding a point at infinity shouldn't turn it into a complete metric space, so that wouldn't be compact

@Perturbative and @Akiva lmao, how's it going?

5:49 AM

So why can't $\Bbb Q$ be an open sunset of a Hausdorff compact set?

@Daminark hiiiiiii

@Daminark I think the problem is that it fails to be Hausdorff

@Akiva maybe something about Baire category theorem?

@Daminark I used to be really good in Mechanics

@Adeek yo, and lmao I was never good at anything physics :P

5:50 AM

today someone asked me a question and I wasn't able to picture it properly :s

@AkivaWeinberger The closure of $\mathbb{Q}$ is $\mathbb{R}$ which isn't compact

you lose something if you don't use it

@Perturbative $\Bbb Q\cap(0,1)$, then (which is homeomorphic to $\Bbb Q$ anyway)

@Daminark It's going aight just getting back into stuff after a break

@Perturbative is that enough to conclude? You can still embed $\mathbb{R}$ into a compact metric space (circle via one-point compactification)

5:51 AM

We want $\Bbb Q$ to be an open set in a Hausdorff compact space

Okay so let's say $X$ is some compact Hausdorff space

or, rather, to prove it impossible

So yeah turns out that the one-point compactification of $\Bbb Q$ is compact, just turns out not to be Hausdorff

I see

So I guess first step is to cover this thing with a billion open sets?

And take a finite subcover and have it be useful somehow

Well my guess would be to show that singletons are nowhere dense in a compact Hausdorff space

5:53 AM

Oh, I see:

Wait no hold on

Oh OK I see again a way to start (don't know if this finishes but let's see)

Fix a point $p$ in $\Bbb Q$, which is an open subset of $X$ by contradiction

$X$ is Hausdorff, so for every point $x\ne p$ in $X$ we can separate it and $p$ with open sets $U_x\ni x$ and $V_p\ni p$

So make our open cover all such $U_x$, along with $\Bbb Q$ itself so that we cover $p$ as well

(which we can 'cause $\Bbb Q$ is open)

And then take a finite subcover.

And now I'm not sure anymore but this definitely feels like something that looks like it could be a proof

But you probably want to look at the finitely many $V_p$-things that correspond to the $U_x$-things?

You know what, you can finish this, I need to go to bed

See you all in the morning