Mathematics - 2023-06-09

anak

2:05 AM

Balarka!

Balarka Sen

Hi @anak!

anak

I came across your papers on stratified spaces the other day and thought of you.

Balarka Sen

Oh lol glad you remembered me

anak

I was trying to find something about symplectic reduction and it turns out you did a stratified symplectic reduction.

Hope you have been well and having fun with math still.

Balarka Sen

Sort of. Our version of stratified symplectic spaces do not quite gel well with symplectic reduction, but we state a relationship.

I was very tired after thinking about stratified spaces for a long time, so I switched gears and now I am trying to learn contact topology.

anak

2:14 AM

Contact topology is thrilling.

I was going to try to do some stuff with contact geometry, but I don't think anything will come out of it. I have been doing mostly just Riemannian stuff recently, so I strayed from my beginnings.

What sort of stuff are you looking at in contact topology?`

Balarka Sen

Oh cool. Maybe we can read together again, finally.

@anak My MSc thesis, which I am currently writing, is about loose Legendrians. I am thinking about Honda-Huang a lot lately.

anak

Have you been in contact with Emmy at all?

Balarka Sen

Nope, but that'd be fantastic lol

Maybe I'll have an occasion someday

I'm a big fan of her work

anak

Yeah, you were really into the Eliashberg-Murphy paper.

Balarka Sen

What are you doing in Riemannian geometry?

(In the sense of, what the hell are you doing over there, chief?)

anak

2:22 AM

flopping around like a fish trying to find an interesting enough result for a thesis and failing

the usual

Balarka Sen

Average grad student life

anak

Pretty much. Found any good music lately?

Balarka Sen

I think I went through a couple of musical phases since we last spoke.

anak

A couple?

Balarka Sen

In the middle I was listening to Burial a lot. Future garage/UK garage/dubstep.

The umbrella term for what I was listening to at that time is "hauntology", you can look that up.

anak

2:28 AM

british deconstruction movement, intriguing

Balarka Sen

yeah lol

anak

UK garage is also british...I am sensing a theme

shintuku

is figuring out whether $I_? = \{h/h' \in K[x,y]_{(x,y)}[1/y]: h \notin \langle y \rangle \frac{K[x,y]}{(x,y)} \}$ is an ideal of $K[x,y]_{(x,y)}[1/y]$ messy or am i doing it wrong

anak

the notation is sadistic in the least

Balarka Sen

@anak indeed

anak

2:36 AM

I think the only british house artist i know well is Fatboy Slim

but he's like big beat/acid

shintuku

$\langle y \rangle$ powers of $y$; $(x,y)$ ideal at which we localize, $[1/y]$ is what we're extending by.

anak

Any current rec.s for what you are listening to right now?

Balarka Sen

I really enjoyed "Engine of Hell" by Emma Ruth Rundle, which I heard a couple weeks ago.

anak

but she isn't from the UK!!!

Balarka Sen

Yeah, I got over the UK phase :P

anak

2:39 AM

American post folk now? Or this is a one off?

Balarka Sen

One off, do not have any coherent themes ongoing.

anak

Sometimes it's good to not have a musical commitment.

Balarka Sen

True

anak

I like those seasons where I can jam to pretty much any tune.

Balarka Sen

2:59 AM

Fill tool is a bit strange in Inkscape but this will do I think

Sourav Ghosh

3:13 AM

Does there exists any set $A$ containing the Cantor set $C$ such that $D(A) =A-A$ doesn't contain any non empty open interval?

shintuku

i used a frac instead of a set minus

we call that optimal notation

Sourav Ghosh

Q.2) Let $A_n\subset \Bbb{R}$ be such that $D(A_n) $ doesn't contain any interval. Can $\cup_{n\in\Bbb N} A_n$ contain an interval?

@shintuku This is not setminus.It is Minkowski difference or algebraic difference.

We call the "difference set"

Balarka Sen

Doesn't $A - A$ contain $C - C$? Isn't $C - C$ an interval?

shintuku

sorry i was talking about a notation mistake i made above, not your comment

Sourav Ghosh

@BalarkaSen Done!

Ignore :)

Both problems are done:)

So $S=\{A\subset \Bbb R : D(A) contains an interval }$ is not a sigma ideal.

Countable sets, measure 0 sets, meager sets, sets having the property of Baire are sigma ideals.

I am trying to find more interesting sigma ideals in \Bbb R

*Correction: $D(A) $ doesn't contain any interval.

Thomas Finley

3:41 AM

What are the various ways/strategies to prove that a set G (G\subset R) is dense in R ?

shintuku

choose a point in R and make an $\epsilon$ neighborhood, and find at least one element of $G$ in that neighborhood

Thomas Finley

Now, I know the elementary defn that a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y

@shintuku is that a definition ?

Balarka Sen

shintuku

it's equivalent @ThomasFinley

Balarka Sen

improved

Sourav Ghosh

3:43 AM

@ThomasFinley cl(G) =R

Any non empty open set intersect G

Thomas Finley

64

Q: Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$

I am having trouble proving the statement: Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.

Say for eg in this post

The user suggests a point in OP to prove the set is dense

It says for every epsilon>0, (0,epsilon)\cap S is non-empty

Is that a criteria for being dense in R ?

I mean an equiv defn or something?

Sourav Ghosh

Side note: Any additive subgroup of $\Bbb R$ is either discrete or dense.

Thomas Finley

I am not quite sure about it?

Sourav Ghosh

G\subset R is dense iff every basic open set ( i.e open interval) intersect G

Thomas Finley

@SouravGhosh yes, but is :"for every epsilon>0, (0,epsilon)\cap S is non-empty" a debn for being dense in R ?

Sourav Ghosh

3:47 AM

No. One way is true.

Thomas Finley

Because this criteria/defn takes only sets (0,\epsilon) into consideration

@SouravGhosh So what about the link of the post I posted here ?

Sourav Ghosh

If $S$ is dense then $S\cap (0, \epsilon) $ non empty for all \epsilon >0

Thomas Finley

@SouravGhosh Oh, but then in the post, the user is trying to prove, S is dense by the other way

Sourav Ghosh

For an example take [0, \infty)

Thomas Finley

That's erroneous, isn't it?

Anyways, I am trying to prove, that the set $\{m+n\sqrt 2:m,n \in\Bbb Z\}$ is dense in R. I dont know where to start ?

Sourav Ghosh

3:51 AM

@ThomasFinley The additive structure will be crucial here.

Thomas Finley

My knowledge about dense sets is that: a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y or equivalently, a set G is dense in R, iff, every point of R, is a limit point of G.

@SouravGhosh You mean that the set is a group under addition?

Sourav Ghosh

$S\cap (0, \epsilon) $ non empty implies $S\cap (-\epsilon, 0) $ also non empty for every $\epsilon>0$

Thomas Finley

@SouravGhosh but why will it be useful here ? Has this something to do with a set being dense in R ?

Sourav Ghosh

See the second answer math.stackexchange.com/a/73272/977780

A clean proof proofwiki.org/wiki/…

AMRIT KUMAR PATEL

4:19 AM

I just got the ability to chat here :)

Thomas Finley

@SouravGhosh this is indeed helpful but here in this case, S is an additive subgroup, so what if it's discrete ?

Koro

4:40 AM

@ThomasFinley Forget m for a moment and consider only the fractional part of $n\sqrt 2$.

then show that it is dense in [0,1).

@ThomasFinley think of discrete as without any limit points.

if you know groups then you can use the fact that additive subgps of R are either dense in R or cyclic.

but that's not necessary really. You can prove this using pigeon hole principle (which is roughly - if you have n>m objects, m placeholders for those objects, then atleast one of placeholders will have more than 1 object in them).

closure of totally bounded set is totally bounded?

cl A= \cup G, where G is open in cl A. Take any e>0. Take a p in cl A.
I want to show that B(p, e) sits inside some G.
B(p,e)= {a\in cl A: d(a,p)<e}
if p is in A \subset cl A, then it does sit inside some G by total boundedness of A so let's suppose that p is in cl A\A.

A\subset \cup G. Given any e>0, for every a\in A, there exists a G such that B(a, 2e)\subset G. closed ball B(a,e)\subset B(a,2e)\subset G for some G.
So B(a,e) (this ball is in cl A) should also be in G.

(first attempt, it should probably work.)

Sourav Ghosh

It looks fine to me.

Koro

thanks.

2

Sourav Ghosh

4:58 AM

$\forall \epsilon >0$ there exists $a_1, a_2, \cdots, a_n\in A$ such that $A\subset \cup_{k=1}^{n} B(a_k, \frac{\epsilon}{2}) $
Then $cl A \subset \cup_{k=1}^{n} B[a_k, \frac{\epsilon}{2}] \subset \cup_{k=1}^{n} B(a_k, \epsilon) $

Koro

I think this is what I should have done.

my answer is not correct. I confused between Lebesgue covering and total boundedness somehow.

I started with cl A = union of some open family

and then tried to find e>0 such for every x in cl A, B(x,e) is in some member of the open family.

this was not required at all.

Thomas Finley

@Koro can you please elaborate the meaning of discrete, i.e some definition? I only know that intutively?

Koro

@SouravGhosh indeed, this is how one would do it. thanks.

@ThomasFinley $A\subset R$ is said to be discrete if $A$ has no limit points in R. e.g. the set of all integers Z

Sourav Ghosh

@ThomasFinley Derived set of A i.e $A'=\emptyset$

Thomas Finley

5:17 AM

@Koro I tried to prove that $n\sqrt 2$ is dense in $[0,1)$, using the fact that irrationals are dense in $\Bbb R$ and so in [0,1) hence, for any x,y in [0,1) we have, an irrational i s.t x<i<y ...but then I am stuck?

Sourav Ghosh

An interesting fact: $Z, √2Z $ both are discrete but their algebraic sum/Minkowski sum Z[√2] is dense.

@ThomasFinley $\{ √2 n : n\in\Bbb Z\}=√2Z $ . Isn't √2Z discrete? Does it have any limit point?

Koro

5:52 AM

@ThomasFinley I meant fractional part of $n\sqrt 2$ not $\sqrt 2 n$. Anyways, here is one simpler way (it doesn't use showing fp of $\sqrt 2 n$ dense in [0,1) first): step 1: show that $0$ is a limit point of {m+ n\sqrt 2: m,n integers}=T. step 2: take any interval (a,b), a<b and show that this interval has an element of T. This uses step 1 and also completeness of R.

shintuku

$A \subseteq B \subseteq C$ with $A,C$ local rings and $B$ a ring does not imply $B$ local

this is a sad moment for $K[x,y]_{(x,y)}[1/y]$

Koro

@ThomasFinley Denoting $\sqrt 2$ by $b$, note that for any integers $n\ne m$, $\{nb \}\ne \{mb\}$ (here {,} denotes fractional part) because if $\{nb \}= \{mb\}$, then $b=\frac{[nb]-[mb]}{n-m}$, which is rational but b is irrational so this is not possible.
Step 1 from the last message: Take any $n\in N$. Divide [0,1) into $n$ parts: [0,1/n), [1/n, 2/n),... Consider n+1 numbers {b},{2b},...,{(n+1)b}. The subintervals are only n but the numbers are n+1 (distinct by the remarks in the first part of this message) so some subinterval must contain at least two of the numbers-say {i b} and {jb}. …

now, you try step 2.

PM 2Ring

@Thomas Dirichlet devised the pigeon-hole principle for precisely this stuff. See math.stackexchange.com/a/56054/207316

It's a beautiful result. And it's probably a good idea to play with a few simple examples until it "clicks".

On a (somewhat) related note, take a look at Beatty sequences. en.wikipedia.org/wiki/Beatty_sequence

robjohn

6:39 AM

@TedShifrin: good evening.

Ted Shifrin

Hi, Mr. sheriff!

robjohn

it was a beautiful day, here. Mid 70s, sunny. There is a chance of rain on Sunday.

Ted Shifrin

6:53 AM

@robjohn Well, maybe here, too.

robjohn

We have a bunny living in our backyard. We just hope it's successful at avoiding the coyotes that come through the neighborhood.

Balarka Sen

7:09 AM

Consider the first Reidemeister move, but take an axis and spin it to get a "Reidemeister move" for 2-knots in R^4 (in terms of knot diagrams, immersed surfaces in R^3).

Iterate to get it for $n$-knots in $\Bbb R^{n+2}$. What should you call this thing? "Torus symmetric Reidemeister move"?

There's different ways to generalize the first Reidemeister moves, hence the question.

Ignore the fact that there are no knotted $S^n$'s in $\Bbb R^{n+2}$ for $n \geq 3$. I am working with knots with more structure, so this is no longer true in my setup.

Torus-symmetric... is a lot of words to say. Toric would be better, if not for that word already being taken. Toral sounds pretentious.

Toroidal? lol

Hopeful Whitepiller

7:49 AM

is the strike till going on

PM 2Ring

8:11 AM

@HopefulWhitepiller Yes. Over 1100 people have signed the letter, and the strike announcement post has over 74k views & +813/-10 votes.

Koro

8:24 AM

@冥王Hades Do you remember Hashirama Senju once said-always conflicts, no matter what the era?

Dannyu NDos

8:40 AM

Has there been a Fields Medal for numerical analysis?

PlaceReporter99

9:12 AM

@PM2Ring I wonder who those 10 downvoters are... (spoiler: probably the staff)

PM 2Ring

@PlaceReporter99 Probably not. There are some huge fans of ChatGPT who are happy to vote against anything that seems anti-ChatGPT to them.

PlaceReporter99

@PM2Ring like the staff...

PM 2Ring

:) No. The staff / company actually don't want the sites flooded with ChatGPT stuff.

PlaceReporter99

@PM2Ring they should step down then and let the community run themselves.

PM 2Ring

But they are freaking out about the drastic reduction in traffic on the network, especially SO, since the release of ChatGPT.

PlaceReporter99

9:19 AM

@PM2Ring It's because most people are asking ChatGPT now.

PM 2Ring

That's certainly an effect.

Numbers have been dropping for years, for a variety of reasons, ChatGPT has just accelerated that.

One of those reasons is that SE Inc seems to be persuing quantity at the expense of quality. That has frustrated and discouraged expert answerers, disappointed in the increasing percentage of low quality questions & answers that have been accumulating, and has contributed to an atmosphere where new members (especially question askers) feel hostility.

So for several years, we've had inflated numbers of people making low-quality contributions. And a lot of those peopke who were asking low quality questions on SE can now ask ChatGPT their crummy questions. ;)

PlaceReporter99

Can’t we just go against the command of SE staff?

PM 2Ring

@PlaceReporter99 Not exactly, but there are ways... See how Physics.SE are dealing with this issue. The Physics mods don't care about ChatGPT, they just use their usual policies on people who persist in posting low quality content.

144

Q: Physics.SE remains a site by humans, for humans

We, the moderators of physics.SE, are deeply concerned about the recently announced "network policy" regarding computer-generated content on the SE network, which effectively attempts to forbid moderators from issuing suspensions or deleting posts for the undisclosed usage of computer-generated c...

discussion featured questions answers policy

That works on Physics.SE, because all the mods are very knowledgeable, so it's not too hard for them to recognise when a post is plausible-looking rubbish. It doesn't work so well on SO, with its much larger volume of traffic covering a broad spectrum of technologies that no single person can have expertise in.

Balarka Sen

9:57 AM

logged into an image ai website and asked it to remove some high tension wires from the foreground of a random image

after many failed attempts, it added more wires

this stuff is so bad

PM 2Ring

Current AI can do some impressive stuff, if you steer it correctly, and are aware of its limitations. But it can easily go off the rails. It's scary that so many people think it's good.

ChatGPT is pretty good with language structures. OTOH, it can't even balance parentheses. See writings.stephenwolfram.com/2023/02/…

Balarka Sen

do i really want to read an article on ChatGPT by the author of A New Kind of Science

PM 2Ring

10:19 AM

@BalarkaSen It's actually a pretty good article. And he (mostly) restrains himself from reminding you how awesome Stephen Wolfram is. ;)

He wants to train a LLM on his writings. He's got decades of material in his archives...

Balarka Sen

lol

apt

from what i can see the greatest contribution of Stephen Wolfram on this planet is helping a couple thousands of precalculus students cheat on their homeworks

which is a feat

and now chat gpt is taking his place, so a collab will be one for the decades

PM 2Ring

From Scott Aaronson's blog, scottaaronson.blog/?p=7188#comment-1948579 > If you’d asked me 20 years ago, “how will Wolfram react to one of the great AI developments of the 21st century?” — I would’ve said, “presumably, by figuring out some way to make it about him.” He doesn’t disappoint! 😀

They developed a plugin to let GPT-4 interact with Wolfram alpha. It still requires manual supervision to correct hallucinations, but it is pretty impressive. writings.stephenwolfram.com/2023/01/…

Koro

11:25 AM

Why is real part of the function $f(z) = z^{1/2}$ positive?

Sourav Ghosh

CheatGPT

s.harp

11:48 AM

@BalarkaSen he's a great expositor and has a knack for pointing out beautiful things of a certain kind

Prithu Biswas

I have no idea why people are so obsessed with ChatGPT as if it is the solution to every problem for humanity.

@Koro I guess you are on your own in your math journey. Good luck o7

s.harp

@PrithuBiswas chapt gpt is something special from a technology / society perspective. This is a new technology that basically overnight became accessible to everybody. And its really impressive. Its natural to think "if this is just the beginning, what will its successors be able to do?". Remember how chess players thought that computer programs will not be able to beat humans in their lifetimes?

Balarka Sen

12:07 PM

@s.harp maybe. i personally would much rather watch a good kung fu movie than read stephen wolfram

im bored with science and tech expositors. they should all jump the AI blackhole and make AIs ghostwrite their books

especially when the expositors are CEOs of multi-million dollar business sharks

Alessandro Codenotti

@BalarkaSen I would much rather watch a bad movie than read wolfram

Balarka Sen

i think when you compare any movie with wolframs writings, it automatically gets the adjective "good" added in front of it

s.harp

I think that these characterisations are coming from a dishonest place. Sure Wolfram is a jerk, a crank, and completely full of himself. So you don't want to give him any credit. But I don't think you can honestly deny that he can write compelling stuff

Balarka Sen

yeah i can

i just did

s.harp

Its the "honestly" that I'm contesting

Balarka Sen

12:14 PM

im being absolutely honest. in fact im so honest that i could never get through 5 pages of anything wolfram has ever written

compelling is in the eye of the beholder, which is fine, really

i think hes boring as shit lol

s.harp

Maybe its my physics background that influences my idea of beautiy then :-)

Balarka Sen

id probably hazard a guess that its just you and not physics

but sure

s.harp

Do you think feed-forward networks are boring?

Balarka Sen

i know epsilon about neural networks

never felt the need to know more than "its just gradient descent on steroids"

s.harp

Yeah, for gradient descent you need to have a candidate for the gradient. With a feed-forward network you get this via is a rule called "backpropagation" which is literally just the chain rule (you have a bunch of layers and the signal goes from one layer to the next, the transform depends on some parameters, if you want to find the gradient you use the chain rule)

from the perspective of mathemnatics everything is barely more htan linear algebra, just a tiny bit of basic analysis

theres a paper from the 80's that is now cited like 20.000 times (number made up) by Hinton where he literally just explains what the chain rule is

still, I think its beautiful, and its not just me. These are the kind of things Wolfram likes as well, if you dont care for them I can understand why you dont enjoy his writings

Balarka Sen

12:23 PM

oh im sure its interesting to many people, i was just suggesting your interests may not be completely decomposed into your physics background

i like physics, a chunk of the math i think about has physical motivations behind it. i admit i dont find neural networks at all interesting

s.harp

You are probably right, this is more computer science aesthetics. But I dont have a computer science background so I cant make a one-liner about it

Balarka Sen

i think interests are complicated emergent social phenomena and thats cool

as long as stephen wolfram doesnt write his complexity theory explanation of interests

s.harp

By the way, does anybody know about this thing mathstodon.xyz ? I stumbled upon it today and have absolutely no idea how to interact with it. Where are the threads / chats?

sunny

1:32 PM

The proof goes on to show that $I(y)=H(y)+C$ by switching the order of integration of $(2)$. Anyway, I am confused about the notation in $(2)$. The partial derivative is taken with respect to $y$, but clearly $f$ depends on $x$ and $t$. On second thought, however, I don't know how I'd write it differently, but it is confusing.

one potato two potato

correct proof is using DCT

sunny

Ok.

It's like writing $\int_a^x f(x) dx$...

The author tries to avoid it, it seems.

Monty

1:56 PM

If you are referencing papers in the middle of a beamer slide show what information do you include? (first author) surname et al, publication year ? Of course full references will be given at the end of the presentation

Astyx

3:01 PM

Depends on the topic, usually I put [AuthorLastNames, year]

If there are too many authors, et al. is acceptable

Xander Henderson

5:02 PM

@TedShifrin In the end, it was good. I made mango salsa (tomatoes, mango, a couple of habaneros, cilantro, lime, salt, red onion), fried up some cod, and made some corn tortillas (I need to get more masa next time I go to the store).

Ted Shifrin

5:17 PM

Sounds yummy, Xander.

Xander Henderson

It turned out well. Though I am not sure if my brother liked it. I wish that he were more communicative. :/

Ted Shifrin

I thought lawyers were plenty communicative when they wanted to be.

Xander Henderson

Though, to be fair, I think that he would eat nothing but roast beef and spaghetti carbonara all day, every day if he had the choice.

@TedShifrin Other brother.

Ted Shifrin

Oh, other brother. Sorry.

Xander Henderson

He has Down Syndrome. I am one of his legal guardians, and he lives with me.

Ted Shifrin

5:18 PM

Right, I only remembered after you corrected me.

Xander Henderson

So I have to find food for him to eat. And he doesn't really like anything green.

Ted Shifrin

At least he doesn't ask for spaghetti-os every day.

Xander Henderson

@TedShifrin I mean, I think if he really had his druthers, he would eat nothing but hot pockets (since, at least, he knows how to use the microwave). But those things are both (a) expensive and (b) so so so salty.

Ted Shifrin

Yes, terribly unhealthy.

I was shocked a few years ago to discover that a can low sodium Progresso soup, which has almost no taste, has something like 1/3 the day’s sodium allotment in one serving.

Xander Henderson

@TedShifrin Yeah. It is frightening.

ZaWarudo

5:41 PM

Let $f:E\to\mathbb{R}$. I know that $f^{-1}(\mathbb{R})=\{x \in E \ | \ f(x) \in \mathbb{R}\}$, is it the same as $f^{-1}(\mathbb{R})=\{x \in E \ | \ \exists a,b \in \mathbb{R} \ \text{s.t.} \ a \le f(x) \le b\}$?

Jakobian

Yes

both are just the set $E$

Ted Shifrin

I don’t see the point of it.

What are you trying to do?

ZaWarudo

Thanks. I want to prove that for $f:E\to\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$ the set $f^{-1}(\mathbb{R})$ is a measurable set given that the sets $E$, $\{x \in E \ | \ f(x)=+\infty\}$ and $\{x \in E \ | \ f(x)=-\infty\}$ are measurable.

We defined $f(x)=+\infty$ if for each $t\in\mathbb{R},f(x)>t$ and $f(x)=-\infty$ if for each $t\in\mathbb{R},f(x)<t$. So I wanted to say that $f^{-1}(\mathbb{R})=(\{x \in E \ | \ f(x)=+\infty\} \cup \{x \in E \ | \ f(x)=-\infty\})^c$.

But I was not sure on how to write $f^{-1}(\mathbb{R})$ with logical symbols, since I never written it explicitly before.

And so I concluded that it should be measurable, because complement of the union of two measurable sets.

Ted Shifrin

5:58 PM

Why not write it as the complement of $\{\pm\infty\}$?

Oh, you said that.

No different from taking the complement of two finite values?

ZaWarudo

No different indeed, sometimes I get scared by new stuff :D

Ted Shifrin

Yeah, we”re not used to functions having actual value $\infty$.

PNDas

6:27 PM

Let $I_K:L^2\to L^2$ and $I_Kf(x)=\int_X K(x,y)f(y)\,dy$ and assume all the required conditions on $X,K$. I want to show that $I_K$ is compact. I know the proof given in Axler's book but I am trying to do in a different manner. Let $(f_n)$ be bounded and $L^2$ is reflexive banach space so it has a weakly convergent subsequence $(f_{n_k})$. Now, $I_K$ can be shown to be continuous so it is weakly continuous. So, $I_K(f_{n_k})$ is a weakly convergent subsequence

Any idea? Is it possible to proceed from here?

copper.hat

7:18 PM

ooohhh, i got a +4 rep bump because a user was removed...

Ted Shifrin

Imagine if they removed me!

PNDas

Last month I got -28 because a question that I answered got removed.

What do they mean by "removed"? I don't understand.

Ted Shifrin

I guess your answer had not been accepted? I think once there's an accepted answer, a question cannot be deleted. Except by mods.

Balarka Sen

7:40 PM

Sanity-check: Say $X, Y \subset Z$ are submanifolds. Saying that $N^*_Z (X \cap Y) \cap T^*Y = N^*_Y (X \cap Y)$ is exactly saying that $X$ and $Y$ are transverse.

I think what I wrote doesn't make sense, but something related does.

I hate thinking dually

Ted Shifrin

Transversality says the conormal bundle of the intersection is the sum of the conormal bundles.

Balarka Sen

$N^*_Z(X \cap Y) = N^*_X(X \cap Y) + N^*_Y(X \cap Y)$?

So just like tangent bundles

Ted Shifrin

No, I meant conormal of $X$ in $Z$ plus ….

Balarka Sen

Ah

Ted Shifrin

but I think what you have is isomorphic ,

Think concretely in terms of $df$ and $dg$.

Balarka Sen

7:55 PM

Shouldn't take long to figure out, but my brain is protesting so I will try to find a different way to say what I want to say.

Yeah, exactly, thinking in terms of functions now

I have an embedding $f : P \to M$, which is transverse to a submanifold $S \subset M$. If $d\phi \in T^*M$ is conormal to $S$ (this just means it vanishes on the tangent spaces to $S$), then $f^* d\phi \in T^*P$ is conormal to $f^{-1}(S)$.

Ted Shifrin

Yes.

skymonkey

8:54 PM

Hi, I apologize if my question is too vague or not well formed. I'm trying to figure out what subject to look at more specifically, or ideally some resources.

I'm interested in the case where I have a differentiable, real valued function f, parametrized by a large set of weights (a good example could be a standard neural network).

I want to consider operators T that act on the function to produce another function of the same type: f' = T(f). I want to use the fixed point theorem to show that some operator is a contraction and therefore applying it many times must lead to a fixed point.

robjohn

9:52 PM

@PNDas The question was unacceptable because it was a PSQ. It was closed and since it was not improved, it got deleted.

leslie townes

10:30 PM

skymonkey: at that level of generality the goal makes sense. but also at that level of generality, it may be no easier to identify which changes in weights produce a contraction than it is to prove that a fixed point exists via some other route.

e.g. if the expectation is that the contraction mapping theorem will make it 'easier' to find/guarantee fixed points, at least in some circumstances, the reality might be, it isn't any easier. it's just translating a hard problem about fixed points to an equally hard problem about what parts of a parameter space yield a contraction.

maybe this would change if you have a lot more structure/hypotheses on the operators involved and the weights.

D.C. the III

10:56 PM

😭😭😭😭😭😭😭😭..........I wanted to graduate next year, but being realistic it isn't feasible.......life is so heartbreaking sometimes........😭😭😭😭

Jakobian

matter of perspective

D.C. the III

Very true and I have been in that frame of mind all day adjusting the perspective

it is coming to terms with it and reconciling it which has been frustrating. Part of the process of things though...

and most importantly being at peace with it

Jakobian

@leslietownes if $U\subseteq X$ is a proper non-empty path-connected open subspace of a path-connected metric space $X$, does there always exist $x\in \partial U$ such that $U\cup \{x\}$ is path-connected?

i.e. I want to find some path contained in $U$ which would connected to the boundary

for $X = \mathbb{R}^n$ for example, the answer is yes

and in general, can have $x\in \partial U$ with $U\cup \{x\}$ not path-connected

Ted Shifrin

@D.C.theIII I often ran into math majors who were taking loads impossible for them to be successful with (like 4 upper-level courses a semester). They insisted they had to do it in order to graduate. I tried to suggest that taking a lighter load and succeeding would be better than failing courses their way.

4

D.C. the III

@TedShifrin That's exactly the reconciling I am coming to grips with........sure I "could" finish and "get the piece of paper", but at the expense of burning out and not learning anything and scraping by with low marks?.............That is exactly how I ended up in the predicament I'm in now.

Ted Shifrin

11:06 PM

Well, pragmatism has different aspects. :)

D.C. the III

In truth the only real "downside" is I will be finishing a year later and I'll be a year older......even if I'm not a complete spring chicken anymore nothing else truly changes. I could still do everything I would like to do....

Ted Shifrin

Sometimes one needs to modify goals …

D.C. the III

it's true.

I'll get through my bout of existential crisis. Just needed to do some digital venting

skymonkey

@leslietownes thank you, I'll think about it

mick

-1

Q: Prime number curiosity : $ \lim_{n \to +\infty} \frac{\exp(1 + \sum_{i=1}^{n}\frac{1}{p_i-1})}{\prod_{i=1}^{n}(1+\frac{1}{p_i-1})} = \frac{79}{25}$ !?

Let $0<n$ be a positive integer. Let $p_n$ be the $n$ th odd prime. $$p_1 = 3,p_2=5,p_3=7,p_4=11,...$$ Let $$f(n) = \exp(1 + \sum_{i=1}^{n}\frac{1}{p_i-1})$$ Let $$g(n) = \prod_{i=1}^{n}(1+\frac{1}{p_i-1})$$ Let $$t(n) = \frac{f(n)}{g(n)}$$ Then this limit exists $$ \lim_{n \to +\infty} t(n) = \f...