Not a hyperplane. A subspace ('cause "hyperplane" means codimension 1)

Or... something I dunno

Someone will be knocking on your door any minute now to take you away...

(the 1960s had some weird songs)

Oh, this dude has a "funny" story about Dirac. He met Dirac once, back in the day (I get the impression that Dirac kind of wanted him to f' off, but was too polite to say so). Do you know what is really funny about Dirac?

He was very tall!

...

No, this is really funny, you know?

Because, like, the Dirac function is very tall!

@Xander why does your advisor invite him? For amusement?

Y'get it?

...ow

Now I need another strange 60s song to relax after that

@Daminark I honestly don't know. I think that my advisor may see it as a kind of mitzvah; dude is clearly out of touch with the world and needs some human contact.

Xander needs to stop storytelling

We don't need to be subject to this cringe

Rip

So what even is the geometric interpretation of the transpose? I know that it inverts rotations, and it leaves things with orthogonal eigenvectors the same

(It's symmetric iff it stretches and squeezes along orthogonal directions, yeah?)

But for other matrices, how do you interpret it geometrically, if at all?

@AkivaWeinberger Meh, it's the dualization, formally speaking

were I any good with the singular value decomposition, now would probably be the time for me to say something intelligent using it

but I'm no good with the geometric interpretation of it

@BalarkaSen Dualization?

Meaning, it's just a thing what reverses the multiplication?

trade a vector for the dual covector, yeah

Ah

The dualization (sending a vector space to it's dual) is a functor on the category of vector spaces over some field. The transpose is the functor's value on the morphisms of the category :3

I am become nlab

Destroyer of math

@Semiclassical SVD is really cool but I forget it a day or two within every time I learn it

same

I'm also spoiled by most examples in physics being either Hermitian or real symmetric

It says that I can decompose a matrix as $U \Sigma V^T$, iirc, where $U$ and $V$ are orthogonal and $\Sigma$ is a diagonal matrix

@BalarkaSen mega same

The point being a matrix eats a sphere and spits an ellipsoid with different coordinate axes

Diagonal means "boxy squishstretch"

but for the standard basis

in which case there's no reason to worry about SVD since you never run into problems with diagnolizability

So decompose that as rotation, boxy squishstretch like @Akiva said, then rotate back again

(as long as the matrix isn't singular)

"Boxy squishstretch" I guess is a conjugation of a diagonal

So it's really just do a rotation and a boxy squishstretch

I get confused on why $U \neq V$ in general

just the box might not be perpendicular to the axes

The amount of rotation to bring stuff in the coordinate basis and to bring stuff back when it's scaled by a diagonal matrix is not the same

@BalarkaSen Doesn't this have to do with a lack of linearly independent eigenvectors?

If they are the same then it's symmetric

Yeah good point @Akiva

and it's a conjugation of a diagonal matrix (aka it's a boxy squishstretch)

or no I'm thinking of diagonalization

Hmm, does that mean a symmetric matrix leaves the axes of the ellipsoid the same?

Nah that'd be diagonal.

that's the SVD of a shear

It's a conjugate of diagonal

Hmm.

Mostly I'm trying to remember if there's a simple geometric interpretation for a symmetric matrix as a linear transformation.

Aka, you can start with some axes on the original sphere (not the coordinate axes), and the symmetric matrix will stretch along those

'Cuz after basechange symmetric matrices are made diagonal

Basechange by orthogonal matrix

Ah, so there's a particular ellipsoid (well, family of ellipsoids) whose axes get preserved

Yep

Makes sense

Whereas for a generic linear transformation there needn't be even that

(and of course there won't be if the matrix isn't square, since it isn't even an ellipsoid of the same dimension after the mapping)

Nobody cares about nonsquare matrices

Only bums do

Eg you

Wow, rude

Square matrices are passé. Pentagonal matrices are all the rage now.

what is a vector but a highly non-square matrix

LOL

@Semiclassical Nice.

Actually, I"ve had reason to care about SVD of non-square matrices lately

but it's annoying and I don't really like it

I saw a really nice talk last year that used SVD to (lossily) compress images.

Hm, can you write the rotation of a matrix 90 degrees in terms of standard operations?

jesus

no

yes. that's just multiplication by the antidiagonal matrix

I cannot

It was kind of cool, and ended up leading into a really nice research project for a group of clever undergrads.

Oh, wait.

Hmm. I know I"ve done it before.

Okay, bedtime. Because it is now at least an hour past when I should have gone to bed.

laters

No, I think you're right @Semiclassical

Isn't it just the transpose composed with a permutation matrix

Permuting left-to-right to right-to-left

no, i'm wrong. It's $A\mapsto JA^\top$

It's not $A\mapsto AJ$?

@BalarkaSen yeah, which is the 'antidiagonal' matrix we were referencing

high fives

no, $A\mapsto AJ$ reverses the order of the columns

Ah

whereas $A\mapsto JA$ reverses the order of the rows

Mhm

And $A\mapsto A^\top J$ is the rotation in the other direction

I had reason to care about this stuff a while back because I had matrices which were J-conjugate to themselves

or maybe it was $A=JA^\top J$? ugh, been a while

So do we live in $\mathbb{R}^3$ or what?

It's within experimental error

Locally and spatially, yes, seems like it :P

A better parsing is that $\Bbb R^3$ is a model of where we live in

yep. one dimension of time, two dimensions of space on the chat room :P

lmao

I mean, technically, special relativity says that it's impossible to separate space and time

at least, not in a way that makes sense for all reference frames

@AkivaWeinberger I don't quite know what that means

Special relativity is a model of the geometry not the topology of spacetime

data expunged

Oh wait no I misread what you wrote

Spacetime is $\Bbb R^4$, yeah, and space is a 3D subspace of that, yeah?

What I was saying is, which 3D subspace it is depends on your reference frame

Locally, at least, yes

It's foliated by the spacelike hypersurfaces

So yes, picking a leaf is what reference frame means

@Slereah Verify pls

What's simultaneous for me might not be simultaneous for you

meaning, two points that are in the same "space" subspace for me might be in different "space" subspaces for you

Or leafs rather than subspaces

Sounds true

You can't separate space and time because Lorentz transformations don't keep them separate, essentially

Right, the Minkowski metric ($ds^2 = dx^2 + dy^2 + dz^2 - dt^2$) doesn't decompose as a direct sum of the spatial metric and the temporal metric.

Weird metric my dude, weird metric

I think it's derived from figuring out what expression is left invariant under Lorentz transformations, and it turns out to be precisely that

Standard physics garbage. Write coordinate-dependent things and then see what doesn't change under switching of coordinates

Somehow that pops out geometric and topological quantities that usually are meaningless and the physicists who discover them bask in the glory

But they're still good invariants so can't complain. Massive fuckers.

It's like a measure of how far inside your lightcone something is, right?

(Is that what it's called, lightcone? I forget)

Yeah, that would be the kernel of $ds^2$

Or, how far outside I mean, 'cause stuff inside is negative

I think the terminology comes from light travelling tangential to the cone

The two lobes of the cone indicate future and past and the axis hypersurface of the cone is the present

@BalarkaSen Picking a frame is more picking a tetrad, usually

Here comes the man

although for Minkowski space, it's generally pretty similar

at least for inertial observers

Is what I am calling a leaf is what you're calling a tetrad

The tetrads of an inertial observer pick a foliation

No

The tetrad is a basis at every point

retracting my pfeh

It's the whole $g_{\mu\nu} = e^a_\mu e^b_\nu \eta_{ab}$ thing

basis that is tangential to the leaf and pointing in the time direction?

so a flowbox for the foliation?

Yeah

Gotcha

For this reason tetrads are also called frame fields

since their basic idea is that they are a local frame

yep makes sense

although of course you can pick Bad Frames even in Minkowski space

then things get tricky

the nice thing with Minkowski inertial frames is that given the frame of a single curve, you can extend it to the entire spacetime

there's a nice correspondance between the "real" coordinates of space and what you can deduce from measurements

not so easy with a more general spacetime

I see. Very interesting

0celo7 once told me you can integrate the cone fields given by the metric to get a "cone" inside the spacetime

i dont know what that means though

The largest $n$ for which a closed form of this value (call it $\Delta(n)$) is known is $n=5$:

$$\Delta(5) = \frac{65}{42}K-\frac{380}{6237}\sqrt{5}+\frac{568}{3465}\sqrt{3}-\frac{4}{189}\p‌​i-\frac{449}{3465}-\frac{73}{63}\sqrt{2}\tan^{-1}(\frac{\sqrt{2}{4})-\frac{184}{1‌​89}\ln2 + \frac{64}{189}\ln(\sqrt{5}+1)+\frac{1}{54}\ln(1+\sqrt{2})+\frac{40}{63}\ln(\sqrt‌​{2}+\sqrt{6})-\frac{5}{28}\pi\ln{1+\sqrt{2})+\frac{52}{63}\pi\ln2+\frac{295}{252}‌​\ln3+\frac{4}{215}\pi^2+\frac{3239}{62370}\sqrt{2}-\frac{8}{21}\sqrt{3}\cot^{-1}(‌​\sqrt{15})-\frac{52}{63}\pi\ln(\sqrt{2}+\sqrt{6})-\frac{5}{7}\alpha+...$$

Wow

ow

$$\Delta(5) =\frac{65}{42}K-\frac{380}{6237}\sqrt{5}+\frac{568}{3465} \sqrt{3}-\frac{4}{189} \p‌​i-\frac{449}{3465}-\frac{73}{63}\sqrt{2}\tan^{-1}(\frac{\sqrt{2}}{4})-\frac{‌​184}{1‌​89}\ln2 + \frac{64}{189}\ln(\sqrt{5}+1)+ \frac{1}{54}\ln(1+\sqrt{2})+\frac{40}{63}\ln(\sqrt‌​{2}+\sqrt{6})-\frac{5} {28}\pi\ln{1+\sqrt{2})+\frac{52}{63}\pi\ln2+\frac{295}{252}‌​\ln3+\frac{4} {215}\pi^2+\frac{3239}{62370}\sqrt{2}-\frac{8}{21}\sqrt{3}\cot^{-1}(‌ ​\sqrt{15})-\frac{52}{63}\pi\ln(\sqrt{2}+\sqrt{6})-\dots$$

Rip in rekt

My chat is getting tilted

@BalarkaSen Do you mean the chronological future, maybe

I do not know the right formalism

Chronological future $I^+(p)$ is the set of points that can be reached by a timelike curve starting at $p$

It's the generalization of the light cone

Ahh

$$\Delta(5) = \frac{65}{42}K-\frac{380}{6237}\sqrt{5}+ \frac{568}{3465}\sqrt{3}-\frac{4}{189} \p‌​i-\frac{449}{3465}-\frac{73}{63}\sqrt{2}\tan^{-1}(\frac{\sqrt{2}}{4})-\frac{‌​184}{1‌​89}\ln2 + \frac{64}{189}\ln(\sqrt{5}+1)+\frac{1}{54}\ln(1+\sqrt{2})+\frac{40}{63}\ln(\sqrt‌​‌​{2}+\sqrt{6})-\frac{5}{28}\pi\ln(1+\sqrt{2})+\frac{52}{63}\pi\ln2+\frac{295}{25‌​2}‌​\ln3+\frac{4}{215}\pi^2+\frac{3239}{62370}\sqrt{2}-\frac{8}{21}\sqrt{3}\cot^{‌​-1}(‌​\sqrt{15})-\frac{52}{63}\pi\ln(\sqrt{2}+\sqrt{6})-\dots$$

Stop

Drink some water

STOP I say

IT COMPILES

put a space

somewhere

It's beyond saving

Good

moving on

I hope you aren't getting into that habit of making predictions based on float evaluations. But ill assume there is an actual derivation of what you are saying that is private

water is pretty good yeah. I like meth too

Wtf

Lmao

well it gets me out of bed

I can’t tell if your joking or not

Coke is pretty good

I like coke too

oh right public image n shit yes sorry I meant I like your method sir

You are who you are , it was just out of nowhere so am assuming it’s a joke

@Akiva the real coke or the fake coke

@Zee It will forever be a mystery

@BalarkaSen The new one they just released

i haven’t seen this kid in a while ^

Who, me?

Yeah I've been gone for a bit, hi

Ya

How's life

It is what it is

but seriously show the derivation for the Catalans thingy if you are going to put it up

how about you ?

It's been good man

what math have you been flirting with

Made some music

dude , hook me up

um I'm in no way familiar with that terminology

@Adam What brings you here

what?

cdn.discordapp.com/attachments/408682268115075073/…

He prob confused meth with math

Does that work?

Tiny thing I made for a weekly competition

@Adam That was meant to be a "Hi". I haven't seen you around in this chat.

@Zee Thing I made a while ago (there's a play button somewhere)

Someday I'll learn how to actually produce and mix stuff

And do instrumentation

oh hello yes I've been on here before. We also know each other from MMF but I cant remember what user name I used back then I just remember there was some prodigy kid from indian with the username Barlarka that I discussed something id done with

Oh uh yeah I was that person. But ignore that version of me.

That's his real name

Ignore that version of me, lmfao

Second one isn’t playing sound

also never allowed to return ever on MMF. I mean unless I spend 5 minutes making an account. which I won't because it's the shittiest website to exist ever

You hit the play button on the top? @Zee

Ya the notes move but no sound

Alternate link

It's a pretty old school forum. I haven't been there in a while

@BalarkaSen Where are the patch notes for Balarka v4.20

I'm not downloading your mp3 just show your derivation

@Alex I think you mean Balarka v3.60 (noscope demo)

Bro

Gottem

TBH I didn’t like the first one but the second one was awesome

Yay ^_^

You played that on piano ?

Nah, I put the notes into the computer :/

Also, a thing I made for my last assignment of high school (which was to arrange a song in four-part vocal harmony)

That’s fine , that’s the future anyway

Akiva u should make vaporwave

what the hell Zee how did you get a badge called Outspoken

(I chose the Hebrew prayer Ana Bekhoakh)

But for real , I really like the second one

Thanks

What badge ?

(The vocal harmony one is also computers)

It’s alright , second one still better

I'm assuming your message was preapproved by a real fraternity member, i.e it was considered to be conveying a socially correct attitude by a man. Meh I lurked on your profile and saw a badge called outspoken. Altruist is still the funniest

We get badges ?

https://math.stackexchange.com/help/badges

You get the badge Outspoken if you have posted 10 messages with a star each by 10 different users

Oh that’s boring

I thought I got it for literally being outspoken

Which you can’t deny I am

I make mouth noises

r/woosh

Well it's more of a psychologically reward based spectrum to encourage users that are quite heavily on the spectrum as they say to post content that increases or maintains the websites monetary value and never question why they should do that for free forever

psychological*

lmao

Are you the Adam from the facebook group >implying we can discuss math, or similar?

and first spectrum was meant to be system

Are you in meth ?

That's not mutually exclusive

Wow , for the whole tome I didn’t realize there was an Alex and an Adam

also not allowed to comment or post in implying until friday

I thought you were the Same avatar

I am basically the poster child for sanituy

sanity\

stupid typing requirement

So anyway Akiva what math are you dabbling in now ?

Not really anything right now tbh

I've been thinking about other things for a bit

I see

Well your ahead of most math students your age so you can def afford that

What have you been thinking about ? You know thinking is dangerous

I need to learn category theory

and I've tried before and didn't really get it

I think I need to learn by doing/playing

which is a problem 'cause I'm lazy

Why do you need to learn it ?

'Cause I've been meaning to for a long time

Couse...?

Bunch of other things use the language of category theory

So....?

So if I want to learn them I should learn it

That’s a leap of logic

Also one day I'm probably gonna be taking category theory in an actual class, and I don't like to take a math class about a subject I don't already know

I know general topology but I never studied set theory

('cause I haven't done so in, like, years?)

And most schools don’t have a catagory theory class

And you should not take classes in things you already know

You won’t be able to anyway

I just graduated high school

I know

I assume there's category theory in universities

Sure there isn’t catagory theory class , maybe in a few but not most

Eh whatever

Seriosuly , I go to Stony Brook , students here don’t learn catagory theory except for a couple weeks in the second core course of algebra

They had a catagory theory seminar once , for undergrads , it was a disaster

Homological algebra is a course though

I was about to say, "Hey, that's not too far from where I am"

and then I realized that Long Island is pretty big

and it's probably about the same distance from me as Yale is

I know , your in sheepshead bay right ?

No

It’s about a couple hours driving from there

I am in Brooklyn

I used to live there

Then I moved here , screw that place , too loud

Anyway , I don’t wanna be the anti catagory theory guy so I’ll let you be

Hi skull

:P

how's life pal?

It's good my man

coolio

$sin(\frac{3\pi}{7}) \in \mathbb T$ is true right?

@Adam That depends on how you define $\mathbb{T}$

Exercise: Prove that, if $\pi$ were rational (of the form $a/b$ for integers $a,b$), then $\int_0^\pi x^n(\pi-x)^n\sin(x)dx$ would be an integer multiple of $n!/b^n$. Divide by $n!/b^n$ and conclude that $\pi$ is not rational.

like it's definitely non algebraic right, when the rational that is multiplied by $\pi$ is greater than 5?>

sorry $ \lt 1/5$

No, the sine or cosine of a rational multiple of $\pi$ is always algebraic @Adam

As in, there always is a polynomial that it is a root of

Write $\sin(7 \cdot 3\pi/7) = 0$ out explicitly

Plug in $x = \sin(3\pi/7)$

(Remember that Galois proved that "being a root of a polynomial" and "being expressed in terms of radicals" are not the same thing, incidentally)

(As an example, the roots of $x^5-x+1=0$ are not expressible in terms of radicals)

@AkivaWeinberger did you go to the brooklyn school of science?

ok show me that Galois proof you just mention

@student No

if you don't mind me asking :^)

That's why I am getting confused I feel like I should be able to find a radical expression or some composition of radicals for all numbers that are the sine and cosine of a rational multiple of pi

sine or cosine lol

@AkivaWeinberger I feel like he's right. Algebraic values of trigonometric functions should be expressible in terms of radicals.

Well technically you could cobble it out of roots of unity

Yeah there's a theorem to that effect

But I forget it's name

Like I am not disputing the Galois proof but it's obviously relevant

which Galois's proof counts as written out of radicals, actually, so you're right

ok I need to know what it's called asap please sen

in Australian culture asap means when you can be fucked finding it so don't take the military meaning

> All trigonometric numbers – sines or cosines of rational multiples of 360° – are algebraic numbers (solutions of polynomial equations with integer coefficients); moreover they may be expressed in terms of radicals of complex numbers; but not all of these are expressible in terms of real radicals.

- T'Wiki

Casus irreduciblis being the relevant point in the last line.

nice ok

> According to Niven's theorem, the only rational values of the sine function for which the argument is a rational number of degrees are 0, 1/2, 1, -1/2, and -1.

@AkivaWeinberger same tbh

That's a fun one

Essentially guarantees that sin(0), sin(30), sin(45), sin(60), and sin(90) are the only "special" values of sine

I had a proof of that but I forget exactly how it goes. I think you assume $\sin(x)=q$ for some rational $q$ and some rational-times-pi $x$, and then analyze a linear recurrence relation for $\sin(nx)$

$\sin(n(x+2))=2\cos(n)\sin(n(x+1))-\sin(x)$

$f_{n+2}=2Cf_{n+1}-f_n$ where $C=\cos(x)$

and now I realize that this proof only works if I do $\cos$ instead of $\sin$

(since the statement is equivalent either way)

$\cos((n+2)x)=2\cos(x)\cos((n+1)x)-\cos(n)$

so it's actually the same exact recurrence relation

$f_{n+2}=2qf_{n+1}-f_n$ where $q:=\cos(x)$

t

sure I thought that might be straight forward by taking the series expansions

Um, trying to remember how this goes

but really man why ever do you want to talk about trigs in terms of degrees don't enable them

I think if $2q$ isn't an integer then the denominator of $f_n$ grows exponentially

which is bad because it has to be zero eventually

Whatever, you can finish the proof yourself if you want to