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12:08 AM
3
Q: What exactly is averaged when doing batch gradient descent?

BenI 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...

 
12:27 AM
just linking answers you wrote 6 months ago 😂
 
And why are we doing that?
 
12:59 AM
 
 
2 hours later…
2:53 AM
arxiv.org/pdf/2010.02766.pdf martin hairer considers stochastic processes with values in the space of $\mathbb{R}$-trees now
crazy stuff
 
3:05 AM
cool
question, can a transformation be classified as stochastic?
I'd like to understand a stochastic version of a certain transformation
 
@geocalc33 there are things such as measure-preserving transformations, but I don't know if that's the kind of transformation you mean
 
@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
 
3:27 AM
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
 
how do I specify the transition probabilities?
 
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
 
I have an initial distribution
 
@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
 
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
 
3:33 AM
In more mathematical terms: for all lattice points $x$, define a distribution $\mathbb{P}_x$ on the state space $S$
 
I know I'm being very naive
 
@geocalc33 there are things called variables
you're free to use them
 
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
 
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
 
4:30 AM
in JEE Maths Zone, 2 days ago, by Annamalai Sriram
user image
Could anyone help me with the question
Answer for Qn 7 is c and for 8 is b
 
5:14 AM
i have a hard time parsing the question.
 
@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.
 
@feynhat thanks, was just typing that as a question!
 
This likely a problem from the notorious J.E.E
 
BLM
5:33 AM
@feynhat yes ;)
 
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.
 
 
3 hours later…
8:23 AM
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?
 
 
3 hours later…
11:05 AM
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?
 
for a graph, what does the notation $d^-(v)$ mean if v is a vertex?
 
11:24 AM
@AlessandroCodenotti Z/p and Z/2p, multiplication by 2?
 
Ah wait I'm getting confused
 
11:56 AM
@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
 
12:21 PM
Hey ..... I am new in this chat ......
 
Welcome :-)
 
I just wanna know that is there any progress in proving the goldbach conjecture
 
Hello all
 
What are all the solutions to $Id=2\log^2(A)$ for $A$ a matrix whose logarithm is defined.
 
12:27 PM
probably not
don't think a lot of people actively think about it
 
@RanjitKumarSarkar Have you read this?
 
12:43 PM
0
Q: matrix equation. Have I found all the solutions?

geocalc33I'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...

 
12:54 PM
@MikeMiller that makes sense, thanks!
 
 
1 hour later…
2:19 PM
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 $[0,1]$

MikeProblem: $\alpha$ 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...

 
@user6232128 this is from 1949 lmao
 
2:53 PM
@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
 
@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
 
i'll have a look soon
 
h e y\
 
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
 
@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.
 
3:01 PM
@BalarkaSen No. Any triangulation of a 4-manifold is automatically PL, known since 2002ish thanks to pioneering work of some Russian.
Oh
 
That says there is such things right
PL = DIFF
in dim 4
 
Yes it does I got it backwards
 
Fantastic.
We should construct such triangulations
 
sup nerds
 
3:02 PM
@AlessandroCodenotti Grisha, not Misha
 
Perelman???
 
Hey Edward
Do you know about the Jordan-Zassenhaus theorem ?
 
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
 
Oh fine yes of course
Nice
I should learn to speak in this link language
 
@Astyx no haha
 
3:05 PM
:(
 
And simplicial complexes which are homology manifolds up to dimension 3 are manifolds
 
@BalarkaSen maybe we should just learn to speak Russian instead
 
@Astyx which one is that?
 
(actually Milman's concentration proof of Dvoretzky's theorem was published in Russian and I couldn't find a translation of the original paper)
 
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
 
3:07 PM
Oh yes that's what I wanted to say above isn't it
You convinced me of this earlier
 
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)
 
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
 
An affine triangulation ought to be a standard triangulation right
I don't see why the argument really needs compactness
 
@Astyx bleh I don't know it
 
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
 
3:09 PM
these are infinite triangulations so some care ought to be taken no?
you cannot achieve transversality so easily
 
too bad
 
Why not
Baire category should say you can
 
@Astyx I'm making progress on my own problem though! So.. bad luck
hahahaha
 
Take x-axis and take the sin(x)/x curve, how do you nudge them a little to make them transverse?
 
lol
 
3:10 PM
The set of translations/rotations so that chosen finite subtriangulations are transverse should be open and dense
 
But anyway just do it on each compact set right
 
@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
 
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
 
We are saying the same thing but we are both too belligerent to admit it
 
Hahah yeah
 
3:13 PM
Apparently it has some use, for instance for proving there are only a finite number of conjugacy classes in GL_n(Z)
 
What? I don't believe that
Nevermind yes I do
 
conjugacy class of finite subgroups of GL_n(Z)*
 
@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
 
Is there a reason you're working on this stuff?
I could be convinced to get into geometric combinatorics
 
3:29 PM
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
 
Bah
You need to do something easy for my rapidly decaying brain
 
We can try to generalize the above for R^4 lol
Hyperbolic combinatorial exotic R^4
 
@Astyx I know this somehow for certain subgroups of SL_n(Z) at least
 
You can also use it to prove the ideal class of number fields are finite
 
acting on the upper half plane heh
are you doing ANT or what
 
3:38 PM
ye
 
shieeet
me too
 
That's why I was hoping you'd have knowledge of it :p
 
alas, I do not
we proved that the class group is finite using the usual method
á Minkowski
 
Geometry ?
Ye
We didn't :p
 
there's some other way you can do it using Idèles
but idk that
 
3:40 PM
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?
 
Or you can use the Mordel-Weil theorem
 
with an elliptic curve isomorphic to the class group or what
 
something something regular projective scheme
 
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
 
@EdwardEvans The ideles form a compact group and surject continuously onto the class group, which is discrete
 
3:45 PM
oof
ty
 
yo check out this link
 
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
 
You should also try to understand why a dedekind domain is a 1-manifold and why a fractional ideal is a vector bundle
 
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
 
Sure
 
3:50 PM
yeah idk anything about this lol
I ditched AG this semester in place of p-adic Hodge theory
 
lol
 
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
 
soundcloud.com/john-zimmerman-960525209/project-ultra click on this link and tell me what you think
 
sounds like 90s command and conquer music
 
90's kids will remember
 
3:57 PM
unit ready
unit lost
 
autobots roll out
 
(i never played it)
 
HARVESTER UNDER ATTACK
 
Lol
 
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
 
4:02 PM
that's actually not bad
 
all my stuff is written in shitty GP7 midi
true underground
/s
 
You should try growling in midi
 
rofl
just 10000% distortion on a distortion guitar voice
 
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
 
> At 21, Ashwin Sah has produced a body of work that senior mathematicians say is nearly unprecedented for a college student.
 
4:16 PM
Cool stuff
 
@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
 
Suppose one uses the acronym r.v. for random variable, then how would one use the acronym for several random variables? r.v.’s?
 
Yeah I'd use that
 
It’s weird though, cause ‘s is usually used in the context of possession.
 
yeah haha
 
4:24 PM
Although in plural it would be...”the random variables’ common domain”, so maybe ‘ in front of the s has another meaning.
 
4:59 PM
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
 
You’re right.
 
What is the degree of splitting field of $x^3-8x-4$ over $\Bbb Q$?
It's irreducible and has three distinct real roots
 
does adjoining one of the roots give you the rest of them?
 
That's the ultimate question
 
too hard for me!
 
5:07 PM
That equation is actually came from $x^4+2x+2$
It' a resolvent cubic
 
@love_sodam What's the discriminant?
 
Trying to find the Galois of that polynomial
Well I didn't compute that
Could I compute discriminant without finding roots?
 
Yes...
You should know how to do that
 
Oh
just a second
 
I was so bad at this stuff
 
5:12 PM
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.
 
1616
Yes I found the formula
 
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.
 
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?
 
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.
 
So it's 6
 
5:24 PM
That's all.
 
What's wrong with the argument via Galois group?
 
Better to understand why that argument works in the first place
 
Because it's like putting the cart before the horse.
This is how you prove the Galois theory result, as Tobias said
 
Well I'm using those known results of Galois group of degree 3,4 polynomials to compute the Galois group of given polynomial
 
Sure, but you ought to understand the basis of those results before using them
 
5:28 PM
Otherwise it becomes like journalism than math.
 
@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 :(
 
Yes I thought you guys meant using S_3 is nonsense in this context
 
5:58 PM
I think they were just saying its pedagogically offensive, not wrong
 
 
1 hour later…
7:16 PM
For once, I take Balarka's side, @love_sodam.
 
7:46 PM
An historical event
 
7:57 PM
@AlessandroCodenotti And don't you forget it!
 

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