they don't matter much

I figure not, but I'm fascinated nonetheless

@Ted i have one but

well, it's just an idea

It just Ts you off, doesn't it, @Fargle?

not an actual function

Every space is metrizable, don't worry @Fargle

I'd be more upset if you making puns weren't, ahem, perfectly normal.

so we have this curve

@Fargle: your :D

@TedShifrin What if I have two stuff like this intersecting?

I have no idea what you're talking about @Balarka. I said smooth curves.

from 0 to 1, the slope is 0

This is really coincidental

@Ted: English ain't that rigid!

from 1 to 2, the slope starts to increase

@TedShifrin You can just smooth out the corners.

from 2 to 3, it stays the same, and then 3 to 4, starts to increase

I am studying Coxeter groups at this very moment which have quite the connection to these @Bala

Oh, hmm.

and so on

the other curve

Maybe that would destroy infinite intersection

@Krijn Ahh. What are those, again?

from 0 to 1, it has an increasing slope, and 1 to 2 the slope stays the same, from 2 to 3, slope increases, 3 to 4 the slope stays same, and so on

@Bala Ehh, reflection groups XXL

so since they keep interchangably increasing slope, they'll intersect like infinitely many times

and the derivative is constantly non-decreasing

@Zach: The OP wanted strictly convex, but you can probably fix this. You're arranging that the two graphs intersect at all the integers.

So I want a picture like this, essentially:

The circle intersects with the boundary of the polygon infinitely many times

But I am not sure if I can make it smooth

So you want to blow air in the aperiogon to smooth it out, @Balarka?

Maybe smooth out the corners by a factor of $1/n$ as you go towards the limit.

Yeah.

@Ted is my solution half-decent? :P

I think there are easier examples of this idea ;P

@Zach: You're not solved yet, @Zach, but it's a good idea, yes.

Anyway @Fargle if you're bored and want to get some familiarity with them it's a common exercise to prove that a if a space is compact and Hausdorff then it's also regular and normal (normal does imply regular, but it's easier to show regular first and then normal)

Fargle, you already took a Munkres course!!

@TedShifrin I know. I'm reviewing because I got a copy of Steen and Seebach.

Ah ... I gave mine away. Be careful. There are occasional errors in there.

I believe I did the exact exercise @Alessandro is talking about, but it's been a while.

hmm, maybe the difference between these two functions could be $\sin{x}$

@Krijn Do they act by isometries on something?

because then they would intersect every $2\pi$

the challenge would be keeping it convex

I'm just complaining that strictly convex should mean that the second derivative is everywhere positive, not just non-negative.

nah, i was just brainstorming :P

anyways, back to this

second derivative everywhere positive?

yup

@Fargle oh, you've definitely seen it already if you went through Munkres, he uses it in the proof of Baire's category theorem

@BalarkaSen Yeah, for example the symmetry group of a regular polytope is a coxeter group

hmm, that's weird

@Krijn Aha, I see.

oh,

Seems like you might be able to make a couple of parabola-like objects that keep outrunning each other as x tends toward either infinity.

so i could just have the slope increase very slowly

instead of staying constant

@BalarkaSen Classify them and you get the same classification as for classical lie algebras

in those integer long breaks

right, @Ted?

Right @Zach. You just have to rig things carefully for the two curves to cross where you want.

@Fargle: If only you could make parabolas intersect infinitely often :P

I was careful to say "-like".

@Krijn that is not completely true, and you know that

Unfortunately I wasn't careful enough to read what @Zach had written and that he had essentially gotten the same idea.

$x^2$, $x^2+\sin x$ but i'm not sure if they're both convex

First you use the wrong part of speech; next you use valley-girl speak. What is going on? :D

@Krijn Interesting.

@Zach: How do you decide?

@TedShifrin Sorry, I'll say "parabolesque" from now on.

oh, second derivative

duh

@BalarkaSen It should be simple lie algebras over $\mathbb C$

Much better, @Fargle. Thank you.

we get $2 - \sin x$

which is always positive

:D

sweet :P

too bad that was an easy problem :P

It actually was an exercise I assigned in Spivak some years ago, and it took me a while to see different solutions.

But someone just posted it as a diff geo question.

is it answered?

Pretty much ... in the comments.

oh :/

whatever, im here for the learning, not the fake internet points

Speaking of which ... get back to that. And I'm gone for now.

" ... And I'm gone gone gone / now I'm older than movies / ... "

Bye @Ted

And everyone else, I'm going to sleep

Night!

@Balarka you should go to mathcamp or something like that

nah

well, alright :P

maybe one day you'll have a theorem named after you

I'm a frog. Not gonna escape my muddy well.

and i'm going to know you as the guy who wanted all those hats on math.SE

You never know, I might take up doing off-off-broadway theaters and writing nonsense poetry as my professional career.

...the rational polynomials are dense in $C[0,1]$?

yeah

That seems utterly remarkable.

Polynomials are dense in that by Stone-Weierstrass. Rational polynomials are dense in polynomials, right?

Are you surprised that the real polynomials are dense?

I'm surprised by both, to be honest.

It stands to reason that they're dense in the space of all polynomials.

Polynomials $f(x) = \sum_{i=0}^n a_i x^i$ such that $a_i = 0$ if $i$ is not prime are also dense.

Oh good God.

(that is, finite sums of $x^p$, $p$ prime)

Stone-Weierstrass is indeed sort of surprising. My favorite proof is in the exposition by Matthew Bond, "Convolutions and the Weierstrass approximation theorem"

It's self contained and I recommend it

@MikeMiller What

:)

Is that special to the set of primes, or are there other well-known classes of integer where that holds?

@BalarkaSen maybe become a hand surgeon and fix my wrist

You just need $\sum_{n \in S} \frac 1n$ to diverge.

This is a theorem of somebody's.

Ohh, right.

I recall that

My mind is thoroughly blown.

what is $S$?

some subset of N

well, it must be infinite, right?

Yes.

wait no

?

it'd still be countable, but it seems like you'd still need infinitely many degrees.

Yes. Finite-dimensional subspaces of Banach spaces are automatically closed.

which you would, since the sum of the reciprocals must diverge, and a finite sum is finite.

You'd definitely need infinitely many degrees.

What was the first theorem you saw that stunned you?

Like, "My goodness, there's no way."

idk

@Fargle $\mathbb Q$ is just as big as $\mathbb N$

call me stupid but i thought it was pretty interesting that all contour integrals encircling a holomorphic subset of $\Bbb C$ is always 0

Then again, it was my second week in uni or so

Your wording there is not correct, though I know what you mean.

that is, any function holomorphic on that subset

Holomorphic functions integrate to 0 around closed contours in their domain, yeah

@Krijn I think this might be my first one too. If not that, then $|\Bbb R| > |\Bbb Q|$.

I remember being pretty surprised by that too

yo my mom bought these meatballs and they're the shit

they're kind of spicy though

Then a relative of mine - a retired physicist - told me that's because it actually measures work of a force acting along that contour

I remember the thing that solidified me wanting to do pure math and not applied was learning about the Weierstrass function.

@BalarkaSen OH

And work without displacement is zero.

That makes too much sense.

ah, very smart

@Fargle It was certainly an ah-hah moment.

In the last semester I had these moments of being flabbergasted quite often

i'm still flabbergasted quite often

because i suck at this

I guess another thing that's surprised me is that compactness is the "right" notion for what it's getting at, but that it's so abstract.

@Fargle I'm still not convinced about why compact is the right term, no matter how much Bala tells me it is

i wish i had someone IRL to talk to about this stuff

my teacher certainly isnt interested in anything i have to say

@Zach, I would have killed for something like that when I was younger. Math doesn't appeal to most people.

That's why college is nice--a good few professors and students love to nerd out--but that's a long way away for you.

I was studying something and I thought "this is trivial". Seconds later I thought "no, wait, this is amazing". Then seconds later back to "this is trivial"

haha

@Krijn That happened for me with Lagrange's Theorem.

"Well duh, cosets." "OH WAIT!" "Oh, wait."

So it's a trivial result that conveys amazing knowledge?

Yep.

First isomorphism theorem was also amazing when I first saw it

And now it's just really really really obvious

@Krijn It still surprises me, but now just on a higher level.

well at least i have you guys

as obnoxious as i may be :P

@Fargle What do you mean?

@Krijn That it holds in a lot of categories. Set, Grp, Ring, to name a few. That every function hides a bijection.

Or I guess more accurately, that every morphism in these categories hides an isomorphism.

@Fargle Set?

category of sets

I remember being annoyed to no end at seeing the lattices of certain groups given in Dummit-Foote looking so similar and then I see that's exactly the fourth isomorphism theorem. Good times.

I have never heard of a version for set, I think

The first time I saw the first isomorphism theorem I didn't get what was the point of it, now whenever I see a surjective morphism I'm already quotienting by its kernel

The image of a set is the quotient by the equivalence relation $a \sim b$ if $f(a) = f(b)$.

@Krijn Define the equivalence relation $R$ such that $xRy$ iff $f(x) = f(y)$, then quotient the set by this relation.

Then any function can be written as a composition of the projection onto its quotient by $R$ (a surjection), the map between this quotient set and the image (a bijection), and the inclusion of the image in the range (an injection).

@MikeMiller Ah yes sure, I was thinking in $x \mapsto 0$ things

@Alessandro Aren't you supposed to be asleep

i.e. if $f: X \rightarrow Y$, then $f = i_{Y} \circ \overline{f} \circ \pi_R$.

aren't you Balarka?

I'm also supposed to work on my sleepmathing skills

I am!

I AM?!

Oh.

@Krijn It's similar. It's just you don't have a kernel, strictly speaking, you just use the relational quotient (which has a lot less overhead).

if we need to find how much paint is needed for a room and we know its lenght, height and width

Au contraire, I fixed my sleep schedule.

@AlessandroCodenotti Fair enough, good point

how can we find how much gallon of paint is needed if its 350 square feet a gallon

I wouldn't call being awake 24/7 a sleeping schedule

@MATHASKER What parts of the room are being painted? Walls, floor, ceiling, some combination?

I have a kid in my math class who likes to tell me $1+1-1+1-\dots = \frac{1}{2}$ and $1 + 2 + 3 + \dots = -\frac{1}{12}$

@ZachHauk Punch him in the face

On second thought

i can't

Don't

the four walls beside the ceiling

i explain to him they're divergent

but he just calls me stupid lol

@ZachHauk Ask him to show you how to derive the Cesaro summation for that.

That should shut him up.

Hey guys - just got a quick question. Say we have the following:

i just ignore him then, and go back to not paying attention to the teacher blab about 30-60-90 triangles

@ZachHauk Steal 1/12 of a dollar from him and then tell him that to get it back he needs to give you a dollar the first day, two dollar the next day, etc.

$X^TX \Sigma X^TX$ where $\Sigma$ is a diagonal matrix $k*I$

@Krijn What's he gonna do, cut a penny in thirds?

It's gotta be exact, this dude sounds like no joke.

Can I just arbitrarily move the $\Sigma$ to the front to get: $\Sigma X^TXX^TX$?

he also says he knows the solution to the riemann hypothesis, but at least he's joking there

Intuitively it makes sense

he's not joking with those sums...

@MATHASKER Sorry, I'm being slow today--do you mean ceiling or no ceiling?

no ceiling

@Fargle let the engineers figure out the implementation details

@Fargle Just steal 3 $\frac{1}{12}$ths, and multiply the daily amounts by 3

(engineer logic)

@MickLH Fair enough.

but my teacher said that we are going to do lenght * height +length * height + width * height+ width* height

@OneRaynyDay Only if $X$ is invertible.

I don't uderstand why we need height if we are just doint area?

@MATHASKER You need the height of the walls.

@Fargle I see. Why is that?

well isn't the height same as the width

because they are rectangular

Numberphile is to blame. Annoying physics dudes with their smug faces.

Intuitively speaking it seems like $X$ must be a normal matrix, such that $AB = BA$ so I can just iteratively shift the $\Sigma$ backwards

@Balarka that's the kid who i taught induction

He's also Indian, and likes to think he's the next Ramanujan

(and that's quoted directly)

We've had someone like that here before

yikes kid got issues

@OneRaynyDay Oh wait, no, I think it works in general, because if $\Sigma = kI$, then $X^TX\Sigma X^TX = X^TX(kI)X^TX = k(X^TXX^TX)$.

$kI$ commutes with every matrix

ugh i havent eaten in like forever

@Krijn Well, a Romanian Ramanujan.

Yeah. Scalar multiples of $I$ commute with everything.

I don't know why but I thought you meant a general diagonal matrix, though you spelled it out right there.

@BalarkaSen Too be fair, that alliteration is gold

Ah I see. That makes a ton of sense - It was only when I had explained that it was $kI$ when I realized

@Fargle shouldn't the height just be the width?

Hey @Mike! How ya doin

I know, right?

Doin alright

@MATHASKER There are two pairs of walls. One is measured by height x length, the other is height x width.

@Fargle Diagonal matrices don't commute with general invertible matrices

@MikeMiller You're right.

The center of GL_n is kI

there are other kids in my school interested in math

I was just being silly.

(Unintentionally)

they like to get way ahead of themselves

kind of like me without @Ted's scoldings :P

@Krijn @Fargle you know what I think the sum holds actually even for the money... I mean we require that both parties attain infinite immortality, but in the process of paying $3n$ dollars on day $n$ to infinity, assuming a finite time period exists between the creation and transfer for the daily $3n$ dollars, eventually the combined interest should exceed the original fraction of a dollar thus it is "paid back"

@Zach: A Tedsmack for every child in America. That's what education needs.

@Fargle Write a letter to trump, he's probably cool with hitting kids right?