Namely, that the sum of divisors $\sum_{d|n}d$ of a given $n$ is bounded above by $\frac12 n(n+1)$

hmm, and Terry Tao gives a simple argument for $n\geq \sum_{d|n}d$.

So yeah, I may have a solution :)

has anyone read Tao's intro analysis book?

I'm sure many have

:P

oh, balls. i misread what Terry said.

@Daminark $(\exists x \in this_room)(readTao(x))$ is that any better?

Just wondering if anyone thought it was any good as an intro book

Lel, I know

And I've heard pretty good things about it

@Semiclassical guess what?

?

im gonna be posting an actual demo of this very soon. indiedb.com/games/block-builder

finally cranking the levels out.

woot woot

neat

Hey, stupid thing. I am asked to find the angle between the hyperbolas $xy = 1$ and $x^2 - y^2 = 1$ at their point of intersection; evidently it expects me to compute it out. But I think there is something clever that can be said.

Look at the substituition $x' = x + y$ and $y' = x - y$.

oh dear

who asked this of you?

This is the linear map given by rotating $\pi/4$ degrees, right?

So the second thing is just the first thing rotated $\pi/4$ degrees. I think that means they intersect at $\pi/2$ degrees, yes?

heh heh heh

hyperbolic geometry

the answer is not as easy as you think

You really need to stop using names you think you know, but you don't really know.

...

are you not performing geometry on a hyperbolic paraboloid right now?

in a roundabout way

i might be wrong but I think they map out to being equivalent in that you're looking at a slice of it

No, I am performing a curved topology on a uniformly unicorn rumpelstiltskin.

@BalarkaSen now you're just making shit up to be a jerk

@BalarkaSen why do you assume I don't know?

I didn't assume that. It's clearly an established fact.

Ok, I am pretty sure what I said works. Nevermind, chat.

@BalarkaSen how is it established. what proof do you have that I don't know hyperbolic geometry?

I have a convincing proof which is correct.

which is?

What is my personal gain in telling you the proof? pay me $500 via paypal and then we can talk.

dude I know fucking hyperbolic geometry so why the heck are you being a dick?

No you don't.

based on what? the fact that my half asleep brain mistook your question as asking about hyperbolic parabolae and making a minor correlation as those sorts of things are hyperbolic planes?

I have a strong basis, but I am not going to disclose it without cash

Ok, I should get back to work now.

dude I know you're trolling so just quit being a dick

Trolling is the only rational conversation one can have with you. Ask anybody else in the chat and they'd agree

@BalarkaSen I wrote a fricking 50 page essay comparing it and spherical and euclidean geometry. So why don't you take your "proof" and shove it where the sun don't shine.

@BalarkaSen true. Everyone hates me

cause im an idiot

@Typhon what was that essay for?

@DavidVarela uuuh. high school.

masters?

nah. I just write too much on a routine basis

@DavidVarela why? Do you think I am lying?

damn, my high school math classes were a joke

id post it but I prefer my anonymity at all costs

@Typhon lol, I don't think you are lying

im sure balarka does knowing them.

:p

im too tired for this anymore

@DavidVarela it was actually college but dont tell anyone that. They all think im in high school. Better that way. It is embarrassing with all the high schoolers being better than me at math. Makes me look bad.

XD

also I told them I failed all my classes that way I wouldn't look like I was boasting. Don't want to get accused of that in here.

:/

its no use comparing yourself to anybody

@DavidVarela true, but that doesn't mean everyone else doesn't routinely compare themselves to you

and make you feel like shit

I got started on math very late, it kind of sucks that I didn't have a good teacher when I was young. But I'm having a lot of fun recovering lost ground

i never really did bad in math

I just don't go outside of the areas I've learned

i prefer to just do them in more depth

everyone has their own approach

meanwhile we got high schoolers doing complex analysis

as long as they find cool math to show everyone I don't mind it

But whatever

im sick of this place for tonight

alright man, rest easy

@AkivaWeinberger I'm not in high school. I'm in college. I lied to impress you and simple art.

Oh I'm not going to bed

time to partay

Which is basically just me mindlessly watching YouTube till I pass out

I don't sleep much anymore

if you are going to watch youtube check out 3blue1brown if you haven't already

Henlo

τ/4 radians, τ/2 radians, 3τ/4 radians, 2π radians

Chat needs to calm the eff down. I'm guilty of this too. Let's just be about math again.

@Fargle agreed

So, anyone wanna talk about cofibers?

Or something?

(Tryna make sure I get them)

What's a cofiber?

insert "I'm glad you asked!" here

"I'm glad you asked!"

I've heard of it referred to as a "mapping cone" in some contexts

Does it have anything to do with fibers ?

Also, hi

Homotopy fibers, yeah

Damn chat has gone while while I was absent

Okay, well, start there--what's a homotopy fiber?

They can be defined independently, and I kinda learned it cofiber first :P

Alright, never mind, lul

But anyway, so the shtick with cofibers is that if $f:X\to Y$, you define $C_f$ to be the disjoint union of $X\times I$ and $Y$, where you pinch $X\times \{1\}$ into a point and let $(x,0) \sim f(x)$

Finally

That's the cofiber of $f$

So, let's say for the sake of a picture, $X$ and $Y$ are lines--I'm kind of picturing a pyramid?

Well, I find the easier thing is to imagine them as circles

You get a cone which is glued onto the image of $Y$

Alright, I think I get it.

So, make a cone over $X$, and then glue each point of the base to wherever the function takes it in $Y$?

Yup

So now, you may ask, why did we define it like so and not otherwise?

"Why did we define it like so and not otherwise?"

Well, I respond, the answer is that this makes $[Cf,Z] \to [Y,Z] \to [X,Y]$ an exact sequence

What do these brackets mean?

Homotopy classes of maps between pre-comma and post-comma

Oh for reference everything is based

Thank you based homotopy

snaps eyyy

Or at least I think that's how the sequence went

Well let's check

@Daminark How are you defining the maps in the sequence?

It should be precomposition

I'm trying to remember with what

oh god what's that

So the first arrow should be, precompose with the inclusion map

heLP

Precompose meaning right or left compose?

I still don't know how left and right work, but so first the inclusion map

So the inclusion goes on the left, yeah

@Daminark Two things: inclusion of what in what, and if it's inclusion first, shouldn't it go on the right?

Inclusion of $Y$ into $C_f$

Remember that $C_f$ is includes $Y$ as a subset, so this makes sense

Right, but wait, how does this get us from $[C_f, Z]$ to $[Y,Z]$?

So given $g: C_f \to Z$, we take the inclusion $i:Y:C_f$ to get the map $g\circ i : Y\to Z$

Oh herp.

Oh wait a second yeah you're right

Wait, no, what you wrote just made sense.

My mind has a connected orientation double cover

Anyhow, yeah the inclusion map goes on the right

Similarly, we can pull it back again using $f:X\to Y$ to get $f\circ g\circ i : X\to Z$

So under those actions, we now wish to show we have an exact sequence

In what sense? I've only ever seen exact sequences when groups are the structure involved

Are these classes groups?

You don't know this a priori

You make the preimage of the basepoint into the kernel

So the idea here is that you want some $k:X\to Z$ to be nullhomotopic iff it's given by $f\circ g\circ i$

This is best explained via a commuting diagram, I wish I was there in person with a chalkboard but we'll make do

i.e. that $k$ is a kernel element, and you show that it has to be in the image of this first map?

Yeah, and remember that our sets involved are mappings, so you're in the kernel if you're nullhomotopic

@Fargle Nah, there are generally bare sets. I can talk about an interesting case where they admit a natural group and even ring structure after Daminark's done.

Oh hey @Balarka!

@Daminark I guess I don't know what's meant by "kernel" in the more general context of category theory

Like, the kernel being the nulhomotopic maps makes sense but

Yeah this is definitely borrowing terminology here

So now, let's say we have some nullhomotopy between $f\circ g$ and $c_{\star}$

Call that homotopy $h$

Okay, just to be clear.

$g$ is a representative of a homotopy class of maps $C_f \rightarrow Z$, in other words, just a continuous map.

But then since $f\circ g: X\to Z$, we can think about $h: X\times I\to Z$ as a map from the cone

Okay this is bad notation on my part

$g:Y\to Z$ now

So we're considering those functions from $X$ to $Z$ which are given by $f\circ (something)$

Basically, $g$ is now taking the role of $g\circ i$

Where $f$ is an arbitrary continuous map from $X$ to $Y$?

But a priori we don't know that $C_f$ has anything to do with it

$f$ is given

Oh hurrrrrrrrrrrrrrrrrrrrrrr

Okay so to be clear

We are given $f:X\to Y$

Okay, so a nulhomotopy from $f \circ g : X \rightarrow Z$ to $c : X \rightarrow Z$ is given by $h : X \times I \rightarrow Z$, and we instead think of $h$ as being $C_f \rightarrow Z$ by doing the appropriate gluing?

Yeah

The reason you can see this is because $X\times I$ is part of the mapping cone

So you can consider your map $\tilde{g} = "h \cup g"$

Given that $C_f = X\times I \cup Y$

Now we ask whether that which I put in parentheses is well-defined

@Daminark Okay you had $g$ and $f$ backwards.

Alright, but what you put in parentheses has to be well-defined precisely because of how you defined $C_f$

That's the exact equivalence relation that makes the map "$h \cup g$" well-defined from $C_f \rightarrow Z$

Right?

Why yes it is, because we know that $h(x,1)$ is the basepoint, so that's good, and that $h(x,0) = g\circ f(x)$, and we know that $(x,0) \sim f(x)$, so $g(f(x))$ and we're good

Yup!

I'm total garbage with notation and order of composition

Me too.

But yeah anyway so that's good

Okay, so what have we proven exactly?

What did you prove? The dual to $X \to Y \to Cf$ is exact?

So what we've done so far is show that the kernel of $f^* : [Y,Z] \to [X,Z]$ is a subset of the image of $i^* : [C_f,Z] \to [Y,Z]$

(I also never remember whether the star is top or bottom)

@TedShifrin Well that's just it. I'm not looking at the intrinsic curvature. I'm well aware that curvature is lost (or never had to begin with) once we map onto the geodesic. I'm looking to use the extrinsic curvature.

Where $f^*(g) = g \circ f$ and $i*(g) = g \circ i$, I assume.

It's right, yeah. You're pulling a map $Y \to Z$ back to $X$ by $f$.

pullback = top star, pushforward = bottom star

Yeah exactly

Okay woo finally sweats

Now, we need to show the reverse inclusion

So now we need to show that every element of the form $g \circ i \circ f$ is nulhomotopic.

Oh god that was the order of composition?

($g : C_f \rightarrow Z$)

I fucked up bad

Oh wait you're using different notation but yeah

Okay we're rolling with that

Yeah, I'm using your original stuff, lol

Just to LOL TROL U

NOOOOOOOOO

Proof: time-reverse big bang event to the cone point of $Cf$. Quod Erat Bigbangium.

But yeah so, the reason why this should be true is because we crushed $X\times \{1\}$ into a point.

Walk me through that if you could.

@TedShifrin Also, you're right. If I don't have a constant vector field, it'll not produce geodesics. So I'll just have integral curves in the general case. I was mistakenly assuming constant vector field. That makes things weirder as I need to calculate the integral curve rather than just look at the geodesic associated with my first $X_p$. I now get why you said I'd need to take derivatives in any direction, not necessarily in the direction of $X_p$.

I'm trying to remember how the details check out

I think the idea might be to show that $g\circ f$ is nullhomotopic?

We are trying to prove $X \stackrel{f}{\to} Y \stackrel{i}{\to} Cf \to Z$ is null?

Right.

Yeah

Well, $i \circ f$ takes $X$ down to the bottom of $Cf$, right?

It's inclusion of $X$ to $X \times I$

Yeah yeah yeah you're good

Oh so just shift it up then?

Hang on, I really need paper.

Mhm :)

Okay gg

It takes $X$ down to the bottom of $CX$; push it upwards

If you take the product of all primes up to the $n$th prime. You always get an even number.

Hello

Blows my mind sometimes.

@Axoren lul

@Axoren Isn't this just true because the product of anything with an even number is even?

given an orthonormal matrix U and a skew matrix A

So if you take the product of the first $n$ primes, you're including 2?

can we say that

UAU^T is also a sckew matrix?

@Daminark More specifically the latter, yeah

@Daminark rly maeks u thibk

(det( U ) = 1)

@user84 So, for reference, they're called orthogonal matrices

@user8469759 By skew do you mean $A^T = -A$?

yes

@user8469759 Well, what's $(UAU^T)^T$?

@Axoren That's orthogonal. I was talking about skew.

nvm whoops

u good friendo

more specifically you can consider a vector $a \in R^3$

So yeah

^

That's what I was getting at.

and the skew matrix is built on such vector using the hat operator

Spoiler alert :P

so is it or not?

@user8469759 What's the hat operator?

say you have

Cools down your head slightly

v = (x,y,z)

the hat operator is

It seems like the hat operator defines a projection of some sort. en.wikipedia.org/wiki/Hat_operator

summons inner @Ted to smack Demonark

(0 -z y; z 0 -x; -y x 0)