Mathematics - 2020-11-30

nbro

00:08

3

Q: What exactly is averaged when doing batch gradient descent?

I have a question about how the averaging works when doing mini-batch gradient descent. I think I now understood the general gradient descent algorithm, but only for online learning. When doing mini-batch gradient descent, do I have to: forward propagate calculate error calculate all gradients...

Mike Miller

00:27

just linking answers you wrote 6 months ago 😂

Ted Shifrin

And why are we doing that?

robjohn

00:59

Not for Ted

user2103480

02:53

arxiv.org/pdf/2010.02766.pdf martin hairer considers stochastic processes with values in the space of $\mathbb{R}$-trees now

crazy stuff

geocalc33

03:05

cool

question, can a transformation be classified as stochastic?

I'd like to understand a stochastic version of a certain transformation

user2103480

@geocalc33 there are things such as measure-preserving transformations, but I don't know if that's the kind of transformation you mean

geocalc33

@user2103480 yeah I'll look into that thanks

at present I'd like to understand a bounded random walk on a lattice in 2d. I understand that sometimes one uses a probability distribution to give the probability that a "particle" will move to the next lattice point

I'm not sure exactly how to set up the construction. About all I have is a probability distribution and a lattice

user2103480

03:27

You can define it as a markov process

in this case, a markov chain with countable state space. You just need to specify the transition probabilities on each lattice point and an initial distribution

geocalc33

how do I specify the transition probabilities?

user2103480

so that you randomly start somewhere on the lattice and then say that the probability of moving to this and that (nearby) state is such and such. Of course, the probabilities must add up to one

geocalc33

I have an initial distribution

user2103480

@geocalc33 you just have to choose how probable it is to jump from one lattice point to the other. Note that this need not be symmetric, its perfectly fine to say that jumping from a to b is more probable than jumping from b to a

geocalc33

but then don't I need another probability distribution to specify the transition probabilities systematically? I mean there are an infinite number of nodes and I don't feel like randomly picking numbers and labeling each lattice point

user2103480

03:33

In more mathematical terms: for all lattice points $x$, define a distribution $\mathbb{P}_x$ on the state space $S$

geocalc33

I know I'm being very naive

user2103480

@geocalc33 there are things called variables

you're free to use them

geocalc33

so I'm going to set it up like this: $f(x)A$ where f(x) is my initial distribution, and A is a matrix specifying the transition probabilities

user2103480

and for a random walk, you can just choose the same distribution for all lattice points, such that the probability of moving up,down,left,right is 1/4 each

Annamalai Sriram

04:30

in JEE Maths Zone, 2 days ago, by Annamalai Sriram

Could anyone help me with the question

Answer for Qn 7 is c and for 8 is b

copper.hat

05:14

i have a hard time parsing the question.

feynhat

@copper.hat You're given two curves in R^2, C1: y=f(x), C2: y = g(x). You draw the tangents to them at points with same x-coordinate, ie., you draw tangent to C1 at (x, f(x)) and tangent to C2 at (x, g(x)), then these tangents will intersect on the y-axis. Similarly, you draw normals to these curves at points with the same x-coordinate intersect on x-axis.

copper.hat

@feynhat thanks, was just typing that as a question!

feynhat

This likely a problem from the notorious J.E.E

BLM

05:33

@feynhat yes ;)

copper.hat

i hate to admit it but i am struggling with it.

my excuse is that i need to get back to proofing my son's college app essays.

user76284

08:23

Can any $n \times n$ square matrix be decomposed into a product of the form $A = \prod_{i \in [n]} (I + B_i)$ where $B_i$ are rank 1 matrices, or something similar? Does this decomposition have a name?

Alessandro Codenotti

11:05

What is a simple example of groups $H\leq G$ and an automorphism $f$ of $H$ that does not extend to an automorphism of $G$? Is there a geometric construction for such an example?

Anush

for a graph, what does the notation $d^-(v)$ mean if v is a vertex?

Mike Miller

11:24

@AlessandroCodenotti Z/p and Z/2p, multiplication by 2?

Alessandro Codenotti

Ah wait I'm getting confused

Mike Miller

11:56

@AlessandroCodenotti No this example works sorry

It extends as multiplication by p+2

How about something with free groups

Take the subgroup of the free group on 2 letters generated by a^2, b^2, and ab, and the automorphisms of this given by cyclically permuting the generators?

If f is an automorphisms of F_2 with f(b^2) = ab then f(b)^2 = ab. But ab does not have a square root. Note that g: F_2 -> Z, the homomorphism sending a to 0 and b to 1, sends ab to 1, and 1 does not have a square root in the additive group Z

Ranjit Kumar Sarkar

12:21

Hey ..... I am new in this chat ......

user 6232128

Welcome :-)

Ranjit Kumar Sarkar

I just wanna know that is there any progress in proving the goldbach conjecture

geocalc33

Hello all

user 6232128

hi

geocalc33

What are all the solutions to $Id=2\log^2(A)$ for $A$ a matrix whose logarithm is defined.

Mike Miller

12:27

probably not

don't think a lot of people actively think about it

user 6232128

@RanjitKumarSarkar Have you read this?

geocalc33

12:43

0

Q: matrix equation. Have I found all the solutions?

I'm trying to solve $Id=s\log^2(A)$ where $Id$ is the identity matrix, $s$ is a natural number, and $A$ is a two-by-two matrix for which the matrix logarithm is defined. I'm getting the solutions: $$A=\begin{pmatrix} e^{\frac{1}{\sqrt{s}}} & 0 \\ 0 & e^{\frac{1}{\sqrt{s}}} \end{pmatrix}.$$ Are t...

linear-algebra matrices matrix-equations

Alessandro Codenotti

12:54

@MikeMiller that makes sense, thanks!

Mike

14:19

Hello, I have this question regarding orthogonal probability measures. Can anyone take a look and help if possible? Thank you. math.stackexchange.com/q/3928794/792125

0

Q: Show that two orthogonal probability measures have the following representations for disjoint intervals on $[0,1]$

Problem: $\alpha$ and $\beta$ are two mutually orthogonal probability measures in the sense that for some Borel subset $E\subset[0,1]$, $\alpha(E)=0$ and $\beta(E)=1$. If $F(x)=\alpha\{[0,x]\}$ and $G(x)=\beta\{[0,x]\}$ for $0<x\leq 1$, show that for any $\varepsilon>0$, there is finite collectio...

real-analysis probability-theory measure-theory

user2103480

@user6232128 this is from 1949 lmao

user 6232128

:P

Balarka Sen

14:53

@MikeMiller Do you think there are affine triangulations of $\Bbb R^4$ for which there is no isomorphic subdivision?

Exotic combinatorial $\Bbb R^4$'s lol

user2103480

@BalarkaSen mathoverflow.net/a/350034/129213 this is a pretty good explanation why pathwise stochastic integration doesn't directly work. could be interesting to you

Balarka Sen

i'll have a look soon

anakhro

h e y\

user2103480

The regularity condition on F(W_t) is needed so that the "integral" over the distributional derivative is defined, right? Since these are elements of the dual if I recall correctly

Rithaniel

@TedShifrin This is actually how I felt when I started coming around this chat. I had been getting too much of an ego from being a medium sized fish in a smaller pond. Seeing a bunch of people who easily outdo me helps me keep perspective.

Mike Miller

15:01

@BalarkaSen No. Any triangulation of a 4-manifold is automatically PL, known since 2002ish thanks to pioneering work of some Russian.

Oh

Alessandro Codenotti

Gromov?

Balarka Sen

That says there is such things right

PL = DIFF

in dim 4

Mike Miller

Yes it does I got it backwards

Balarka Sen

Fantastic.

We should construct such triangulations

Edward Evans

sup nerds

Mike Miller

15:02

@AlessandroCodenotti Grisha, not Misha

Balarka Sen

Perelman???

Astyx

Hey Edward

Do you know about the Jordan-Zassenhaus theorem ?

Mike Miller

I'm just being intentionally obfuscatory. This is equivalent to the Standardness of homotopy 3-spheres, as the link of any vertex is a simply connected homology manifold with the homology of a sphere

Balarka Sen

Oh fine yes of course

Nice

I should learn to speak in this link language

Edward Evans

@Astyx no haha

Astyx

15:05

:(

Mike Miller

And simplicial complexes which are homology manifolds up to dimension 3 are manifolds

Alessandro Codenotti

@BalarkaSen maybe we should just learn to speak Russian instead

Edward Evans

@Astyx which one is that?

Alessandro Codenotti

(actually Milman's concentration proof of Dvoretzky's theorem was published in Russian and I couldn't find a translation of the original paper)

Balarka Sen

Maybe triangulations on R^4 which gives rise to exotic PL (=DIFF) structures will be terrible-looking wrt the Euclidean smooth structure though

Maybe affine guys really are standard

Mike Miller

15:07

Oh yes that's what I wanted to say above isn't it

You convinced me of this earlier

Astyx

If you have a semi simple Q-algebra of finite dimension A, O a Z-order on A, and M_A an A-module of finite type. Then there is a finite number of O-modules M such that $M\oplus Q = M_A$ (up to iso)

Balarka Sen

yeah that makes sense, exhaust by compact sets, and on each compact set you can achieve transversality -- so on each compact set the PL structures are equivalent

Mike Miller

An affine triangulation ought to be a standard triangulation right

I don't see why the argument really needs compactness

Edward Evans

@Astyx bleh I don't know it

Mike Miller

You can find a common refinement into a "simplicial cell complex" whose cells are polyhedra by intersecting the two triangulations, ever so slightly translated or rotated

Balarka Sen

15:09

these are infinite triangulations so some care ought to be taken no?

you cannot achieve transversality so easily

Astyx

too bad

Mike Miller

Why not

Baire category should say you can

Edward Evans

@Astyx I'm making progress on my own problem though! So.. bad luck

hahahaha

Balarka Sen

Take x-axis and take the sin(x)/x curve, how do you nudge them a little to make them transverse?

Astyx

lol

Mike Miller

15:10

The set of translations/rotations so that chosen finite subtriangulations are transverse should be open and dense

Balarka Sen

But anyway just do it on each compact set right

Mike Miller

@BalarkaSen Just move the x-axis up by epsilon lol

There's only a countable set of epsilon that won't work

The epsilon so that [sin(x)/x]'(epsilon) = 0

aka horizontal tangencies

Balarka Sen

Right, but you have a million many such sine curves maybes. You're saying countably many polyhedra, so no issue

But just do compact sets, no issue either haha

Mike Miller

We are saying the same thing but we are both too belligerent to admit it

Balarka Sen

Hahah yeah

Astyx

15:13

Apparently it has some use, for instance for proving there are only a finite number of conjugacy classes in GL_n(Z)

Mike Miller

What? I don't believe that

Nevermind yes I do

Astyx

conjugacy class of finite subgroups of GL_n(Z)*

Balarka Sen

@MikeMiller Call a triangulation of R^2 hyperbolic if number of faces meeting at each vertex is at least 7

Any two hyperbolic triangulations of R^2 have an isomorphic refinement

Mike Miller

Is there a reason you're working on this stuff?

I could be convinced to get into geometric combinatorics

Balarka Sen

15:29

Oh just because of Riemann mapping theorem. The above is a corollary; fix a triangulation T of R^2, choose a Riemannian metric g such that T becomes a geodesic triangulation. You can scale so that the curvature is in $[-1, -1 - \varepsilon]$. On the other hand impose R^2 with the hyperbolic metric and consider the geodesic triangulation with the same vertices. You have a conformal diffeo R^2 -> H, which has ||Df|| <= 1 and $\|Df^{-1}\| \leq 1 + \sqrt{\varepsilon}$ by Schwarz-Ahlfors-Pick

That means you can nudge $f$ a little bit so that it's simplicial upon barycentric subdivision, by simplicial approximation. Nudging a diffeo a little bit will keep it a diffeo

So there you have it

Mike Miller

Bah

You need to do something easy for my rapidly decaying brain

Balarka Sen

We can try to generalize the above for R^4 lol

Hyperbolic combinatorial exotic R^4

Edward Evans

@Astyx I know this somehow for certain subgroups of SL_n(Z) at least

Astyx

You can also use it to prove the ideal class of number fields are finite

Edward Evans

acting on the upper half plane heh

are you doing ANT or what

Astyx

15:38

ye

Edward Evans

shieeet

me too

Astyx

That's why I was hoping you'd have knowledge of it :p

Edward Evans

alas, I do not

we proved that the class group is finite using the usual method

á Minkowski

Astyx

Geometry ?

Ye

We didn't :p

Edward Evans

there's some other way you can do it using Idèles

but idk that

aderchox

15:40

If a coordinate is in a vector space other than the one with standard basis vectors(Let's call it V"), why does transforming the coordinate using a matrix its columns which are the basis vectors of V" return the coordinate back to the standard vector space?

Astyx

Or you can use the Mordel-Weil theorem

Edward Evans

with an elliptic curve isomorphic to the class group or what

Astyx

something something regular projective scheme

Edward Evans

The wikipedia page on the Adèle ring is the most complete wiki page I've ever seen btw

also

something something Picard group something something class group

Mike Miller

@EdwardEvans The ideles form a compact group and surject continuously onto the class group, which is discrete

Edward Evans

15:45

oof

ty

geocalc33

yo check out this link

Edward Evans

I should probably know why the Picard group is the analogue of the class group in the function field analogy

but I absolutely do not

Mike Miller

You should also try to understand why a dedekind domain is a 1-manifold and why a fractional ideal is a vector bundle

Edward Evans

eh

I'll look at the end of chapter 1 in Neukirch lol

Neukirch says Pic(o) = Chow^1(o) for o a Dedekind domain

Mike Miller

Sure

Edward Evans

15:50

yeah idk anything about this lol

I ditched AG this semester in place of p-adic Hodge theory

Mike Miller

lol

Edward Evans

was possibly a bad idea but the prof taking AG this semester is a category theory nut so I chickened out and took the other course instead

I guess any course on AG is gonna be category heavy

geocalc33

soundcloud.com/john-zimmerman-960525209/project-ultra click on this link and tell me what you think

Edward Evans

sounds like 90s command and conquer music

Astyx

90's kids will remember

Edward Evans

15:57

unit ready

unit lost

geocalc33

autobots roll out

Astyx

(i never played it)

Edward Evans

HARVESTER UNDER ATTACK

Balarka Sen

Lol

Edward Evans

shoots one bullet at enemy harvester

entire enemy force runs onto tiberium field and attacks your single minigunner

the ai wasn't so sophisticated

@geocalc33 I'ma plug too soundcloud.com/edmate/azure-winds

geocalc33

16:02

that's actually not bad

Edward Evans

all my stuff is written in shitty GP7 midi

true underground

/s

Astyx

You should try growling in midi

Edward Evans

rofl

just 10000% distortion on a distortion guitar voice

geocalc33

I have yet to understand how vocals can be engineered to sound almost flawless

thanks Fourier

the things they can do in the stu these days

CowperKettle

Undergraduate Math Student Pushes Frontier of Graph Theory

> At 21, Ashwin Sah has produced a body of work that senior mathematicians say is nearly unprecedented for a college student.

Balarka Sen

16:16

Cool stuff

Lukas Heger

@EdwardEvans the approach using idèles is actually kind of nice. You show that the class group is a quotient of the idéle class group by an open subgroup, which implies that it is discrete and compact, hence finite

schn

Suppose one uses the acronym r.v. for random variable, then how would one use the acronym for several random variables? r.v.’s?

Balarka Sen

Yeah I'd use that

schn

It’s weird though, cause ‘s is usually used in the context of possession.

Balarka Sen

yeah haha

schn

16:24

Although in plural it would be...”the random variables’ common domain”, so maybe ‘ in front of the s has another meaning.

Mike Miller

16:59

The point of grammar rules is to make something intelligible in a mostly unique way for most speakers

If your use of 's is clear to the reader it doesn't matter that it seems nongrammatical

schn

You’re right.

love_sodam

What is the degree of splitting field of $x^3-8x-4$ over $\Bbb Q$?

It's irreducible and has three distinct real roots

Mike Miller

does adjoining one of the roots give you the rest of them?

love_sodam

That's the ultimate question

Mike Miller

too hard for me!

love_sodam

17:07

That equation is actually came from $x^4+2x+2$

It' a resolvent cubic

Balarka Sen

@love_sodam What's the discriminant?

love_sodam

Trying to find the Galois of that polynomial

Well I didn't compute that

Could I compute discriminant without finding roots?

Balarka Sen

Yes...

You should know how to do that

love_sodam

Oh

just a second

Mike Miller

I was so bad at this stuff

Balarka Sen

17:12

I'll tell you the formula. $x^3 + ax + b$ has squared discriminant $- 4a^3 - 27 b^2$. Compute the discriminant and tell me why I asked you to compute it in the first place.

love_sodam

1616

Yes I found the formula

Balarka Sen

And why does this help?

@love_sodam Careful, this is the square of the discriminant.

Or maybe that's what is called the discriminant. Never mind.

love_sodam

Umm $\sqrt{1616}$ is not a rational number so...

Galois group of that resolvent is S_3

degree of that splitting field is so that 6?

Balarka Sen

No need to argue via the Galois group. Why are you quoting results?

You proved $(\alpha_1 - \alpha_2)(\alpha_2 - \alpha_3)(\alpha_3 - \alpha_1) = \sqrt{1616}$

Where $\alpha_i$ are the roots of the polynomial.

Certainly that expression is contained in the splitting field. So the splitting field contains $\sqrt{1616}$

So $\Bbb Q(\sqrt{1616})$, a degree $2$ field extension of $\Bbb Q$, is contained in the splitting field. $2$ divides order of the splitting field. Also, $3$ divides order the splitting field because the polynomial is irreducible.

love_sodam

So it's 6

Balarka Sen

17:24

That's all.

love_sodam

What's wrong with the argument via Galois group?

Tobias Kildetoft

Better to understand why that argument works in the first place

Balarka Sen

Because it's like putting the cart before the horse.

This is how you prove the Galois theory result, as Tobias said

love_sodam

Well I'm using those known results of Galois group of degree 3,4 polynomials to compute the Galois group of given polynomial

Tobias Kildetoft

Sure, but you ought to understand the basis of those results before using them

Balarka Sen

17:28

Otherwise it becomes like journalism than math.

Edward Evans

@LukasHeger Nice :) I'm interested to see if Vogel will run a course on global CFT next semester, we've done so much local junk over the past two semesters but I wanna do global CFT :(

love_sodam

Yes I thought you guys meant using S_3 is nonsense in this context

Mike Miller

17:58

I think they were just saying its pedagogically offensive, not wrong

Ted Shifrin

19:16

For once, I take Balarka's side, @love_sodam.

Alessandro Codenotti

19:46

An historical event

Ted Shifrin

19:57

@AlessandroCodenotti And don't you forget it!

Transcript for

$Mathematics$ Mathematics