Mathematics - 2020-01-23

Thorgott

00:12

Are you asking for a concrete procedure?

Akiva Weinberger

00:37

An algorithm

Tay

01:18

Hey! If I'm reading a paper that has an intermediate result that I can't understand, where can I ask for clarification? Would it be fine to ask on the main site if I put a relevant excerpt from the paper and also include a link? Or is it okay to ask in chat?

Thorgott

01:29

Both are fine. You can ask pretty much anything here and if no one here can help, you can post a question on main, which will be fine as long as you include the relevant context.

Tay

Thanks!

Here's the paper. I'm confused about the proof of the Theorem 25. On page 18 there's an inequality (63), which if you consider the inequality (61), should have the coefficient 3/2 and not 3 in front

This seems basic, but I can't understand why the author would pick a larger coefficient than required

And no amount of context helps

DRP

02:27

Is there a way I can delete a question of mine here?

After a while just shows flag for moderator

anakhro

03:26

Unless it is profane, I don't see why you'd want to delete it.

user76284

04:11

en.wikipedia.org/wiki/Graphon generalizes immediately to directed graphs if the graphon is asymmetric, right?

Akiva Weinberger

04:29

Feb 10 '17 at 21:45, by Akiva Weinberger

Graph theory game http://treksit.com/?thegame

Semiclassical

05:12

@AkivaWeinberger so basically planarity.net but with background effects

Semiclassical

05:30

also, the bonus area is creepy af

Mary Star

06:06

ok, for that we have:

Let $\epsilon>0$. We want to show that there is a $\delta>0$ s.t. if $0<|x-0|<\delta$ then $|f(x)-1|<\epsilon$.

We have that $\left |f(x)-1\right |=\left |\frac{x}{x}-1\right |=\left |\frac{x-1}{x}\right |=\frac{|x-1|}{|x|}$. How can we continue?

Shootforthemoon

06:29

@Semiclassical I see, thanks. No, the question was a little different from the other, since I did not ask for mathematical proofs but for an intuition (also physical) of what was going on. I see that with the system of expansions everything fits well, but I cannot explain that in terms of lengths for example, and I imagine like 0.999... is just a point immediately before 1.

@Semiclassical And also my problem is about the limit XD 1 divided by ten infinitely many times will tend to zero, but it's not zero (imo)

The limit of that procedure is zero, but since whenever we divide we get still a smaller quantity, if I imagine the steps repeated all over the time, we will never reach zero, but something like 0+

Or maybe @Sem we could say in terms of measures smthg like no points can be found between 0.9999... and 1, hence the two coincide?

Shootforthemoon

06:52

But so, if we could also represent in some way the point immediately before 0.999... , that is "0.999...8", then this must coincide as well with 0.999... We could repeat (via an imaginary limit) such procedure for any point, and conclude that every point is the same.

Akiva Weinberger

@Semiclassical Adding to my reading list: arxiv.org/abs/1308.0066

Here's a paradox

which I shared before, a while ago

A finite game is a game which cannot last an infinity number of steps

(Assume two players)

The Master Game is defined as follows: Player 1 suggests a finite game, and then they play it with Player 2 going first

Is the Master Game a finite game?

On the one hand, yes: because they select a finite game, it will eventually end.

On the other hand, no: If it were a finite game, then we can play like this: Player 1 suggests they play the Master Game. Player 2 suggests they play the Master Game. Player 1 suggests they play the Master Game. etc. This can last forever, contradicting our assumption that the Master Game is a finite game.

So, is the Master Game a finite game or not? Where's the flaw in the reasoning?

Rithaniel

07:09

Anyone familiar with proving the Nine Lemma via diagram chase? I'm doing the case when the top two rows are exact and trying to show that the bottom row is exact.

I'm having some difficulty showing that the bottom row is exact in the middle.

I've shown that the image of the injective side is contained in the kernel of the onto side, but the other containment is giving me a headache

feynhat

07:40

@Rithaniel So, you have shown that the last row is a chain complex.

If you think of each row as a chain complex, you get a short exact sequence of chain complexes.

Rithaniel

Oh, that's meta

Thorgott

10:14

@MaryStar you should rethink the "$|\frac{x}{x}-1|=|\frac{x-1}{x}|$"

Edward Evans

10:45

0

Q: Submodule of $R$-Invariants for a Hecke Pair $(R,S)$

I'll just define a Hecke pair here for completeness: Definition: Let $S$ be a monoid and $R$ a group contained in $S$. We call the pair $(R,S)$ a Hecke pair if every $R$-double coset of $S$ is a disjoint union of finitely many $R$-left cosets of $S$. As an example, we have $(\operatorname{S...

Shootforthemoon

11:28

Hi, I'm trying to prove by comparison the convergence of generalized harmonic series, of the form $$\sum_{n=1}^{\infty} \frac{1}{n^{a+1}}$$ where $a>0$. I proved this for $a=1$, showing that $$\sum_{k=1}^{n} \frac{1}{k^2}<1+\sum_{k=2}^{n} \frac{1}{k(k-1)}$$
but I'm not succeeding in generalize this for any $a>1. $

Any suggestion?

Alessandro Codenotti

11:46

@EdwardEvans suggest me some cool metal

Edward Evans

12:02

@AlessandroCodenotti what kind of Metal? Hahaha

Or rather, do you most often listen to metal? So I know not to recommend some obvious bands lol

Alessandro Codenotti

I'd say I listen mostly to prog rock and post rock, but I also listen to a lot of metal

@EdwardEvans Something heavy but technical maybe

Something that Balarka would like let's say lol

Lately I've been getting into Cult of Luna for example

Edward Evans

12:22

Lol fair, I listen mostly to black metal, but do you know The Zenith Passage?

@AlessandroCodenotti the Song Deus Deceptor is cool af

Alessandro Codenotti

@EdwardEvans no, I'll check them out, thanks!

I'm not very familiar with black metal but if you have cool stuff to suggest I'm always looking forward to listen to something new

Edward Evans

Nice one, also Fallujah are great, check out Sapphire or Chemical cave from that album

Alessandro Codenotti

I will, thanks!

Edward Evans

And if you want some „post black metal“ which is more accessible then Numenorean are incredible, and møl

Hehe

Alessandro Codenotti

I know Numenorean, I listen to the last album when it came out and liked it

Didn't expect to see them mentioned here!

Edward Evans

12:40

Yeah Numenorean are great, they played in Mainz recently

Alessandro Codenotti

Nice

There's a lot of interesting concerts in Köln, but they always happen while I'm in Italy somehow

There's a group from NRW that I really like, I missed them last year when they played in Bochum and I'll miss them again in April when they'll play in Köln, both times because I'm in Italy

But I'll go to a Snarky Puppy concert while in Italy in April so that's fine :P

Edward Evans

Nice, I’m seeing Helrunar tomorrow

Loka Lögsaga is Great

But black black metal lol

Edward Evans

14:32

Yo @Lukas, I have a "general" Hecke theory question for ya if you're here :P

Lukas Heger

okay

Edward Evans

So suppose $(R,S)$ is a Hecke pair. Then an element $\sum_{s \in R\setminus S}a_s Rs \in \mathcal{L}(R,S)$ is $R$-right invariant iff for all $s, t$ such that $RsR = RtR$ the coefficients $a_s, a_t \in \Bbb Z$ are equal. Why then does

$$\sum_{s \in R\setminus S}a_s Rs = \sum_{s \in R\setminus S / R}\sum_{\substack{s_j \in R\setminus S\\ Rs_j \subseteq RsR}}a_sRs_j$$

lol

Lukas Heger

well, $R$ acts from the right on $R\backslash S$ and this yields a decomposition of $R\backslash S$ as a disjoint union of double cosets

Edward Evans

Ahhhhhhhh so I'm thinking of $R\setminus S / R$ as "$(R\setminus S)/R$" lol

Lukas Heger

yes exactly that

Edward Evans

14:41

omfg

hahaha

Lukas Heger

this gives that every left coset is contained in a unique double coset

Edward Evans

okay makes sense, thanks :P

Lukas Heger

and so you get that equality

Edward Evans

thanks man :)

Stupid Questions Inc

15:40

Hi chat! In Spivak's Calculus on Manifolds he says: A set with only finitely many points is obviously compact and so is the infinite set $A$ which contains $0$ and the numbers $1/n$ for all integers $n$ (reason: if $\mathcal O$ is a cover then $0 \in U$ for some open set $U$ in $\mathcal O$ there are only finitnely many other points of $A$ not in $U$, each requiting at most one more open set

Now I get the gist of it, I'm just interested in how to formally prove that the fact that $0\in U$ for some open set $U$ in $\mathcal O$ implies that "there are only finitely many other points of $A$ not in $U$"

Alessandro Codenotti

That's the definition of convergence for sequences. $x_n\to x$ iff every neighbourhood of $x$ contains all except at most finitely many of the $x_n$

For a more explicit approach an open $U$ containing $0$ also contains an interval $(-a,a)$ around $U$. Let $m$ be the least integer with $1/m<a$. Then $1/n$ can be outside of $U$ only if $n<m$, and there's finitely many such $n$

Stupid Questions Inc

that gets it done, thanks @AlessandroCodenotti

Thorgott

why is that like the standard example for compactness

Alessandro Codenotti

15:57

I guess because it's the easiest example after a closed interval

Thorgott

16:12

I guess so. I think the compact interval is a cooler example though; it's not immediately obvious and equips one with an important idea.

Stupid Questions Inc

16:50

I don't get why the "since $\alpha$ is the least upper bound of $A$ there's an $x$ in that interval such that $x\in A$" part is true

(2nd paragraph)

Thorgott

What would happen if there were no such $x$?

Stupid Questions Inc

17:09

@Thorgott then $\sup A = a$

Thorgott

$\alpha=\sup A$ by definition in either case

What are the properties uniquely defining $\alpha$?

Stupid Questions Inc

17:37

@Thorgott if $x\in A$ then $x\leq \alpha$

but how do we know that there's some $x\in A$ and $x\neq a$

Ted Shifrin

@Alessandro: What about Snarky Kitty?

Alessandro Codenotti

Snarky Puppy are a jazz fusion (with other influences) jam band that I really like. They'll play in Italy in April and I'm going to their concert

Thorgott

@StupidQuestions that's the definition of an upper bound, but the supremum is a more specific one among these

Ted Shifrin

I figured there should be a snarky kitty to go with it, @Alessandro :P

Stupid Questions Inc

@Thorgott yes, if $b$ is an upper bound to $A$ then $\alpha\leq b$

Ted Shifrin

17:43

@StupidQuestionsInc: I haven't been following your discussion, so forgive the interruption. But if $A=\{a\}$, then of course $\sup A = a$ and there's nobody smaller in $A$.

Stupid Questions Inc

@TedShifrin yes, the non-trivial part that I'm not getting is why there should be an $x\in A\setminus\{a\}$ in the first place

Thorgott

@StupidQuestion right, so if no $x$ in an interval to the left of $\alpha$ is in $A$, then..

Stupid Questions Inc

$\alpha=\sup A =a$

Thorgott

No, how did you get that?

Ted Shifrin

You're talking about Spivak's proof of Theorem 1-3?

Stupid Questions Inc

17:45

well since there's no $x$ to the left of $\sup A$ then that forces $A$ to be as small as possible, i.e. $A=\{a\}$

@TedShifrin exactly

Ted Shifrin

Well, look at the definition of the set $A$.

Oh, and it's important to write $\alpha$, not $a$, here.

Note that the set $A$ certainly contains the point $a$, and we know $\alpha>a$ (why?).

Thorgott

No, we're not talking about all possible $x$ to the left of $\alpha$. We already chose a small interval around $\alpha$ and we are no talking about the $x$ that lie in that interval and to the left of $\alpha$. If none of these were in $A$, then..

Stupid Questions Inc

@TedShifrin that's exactly the part I'm missing

I don't see why $\alpha > a$

Ted Shifrin

Because of the definition of the set $A$. Whatever the element of $\mathscr O$ is that covers $a$ must cover an entire interval with things bigger than $a$. So they're all in $A$.

(I had to get out the book to find the definition of the set $A$, since you didn't include it in your picture.)

Hi @Mats

Mats Granvik

Hi Ted.

Stupid Questions Inc

17:49

@Thorgott but in the definition of $A$ we require that $[a,x]$ has a finite covering, I don't see any a priori reasons for why there should be any other $x$ that satisfies that other than $a$ (apologies for not including the picture of the preceding page)

Ted Shifrin

I just told you that!

Stupid Questions Inc

@TedShifrin let me take a rest and then come back to it maybe I will get an aha moment then

Ted Shifrin

It's an open cover, so any open set (which is made up of open intervals, by the way) containing $a$ must contain a whole interval to its right.

Stupid Questions Inc

@TedShifrin oooh, so the fact that $[a,a]=\{a\}$ has a finite cover means that there's some open interval containing it

Thorgott

That's different from the question you initially posted

Edward Evans

17:51

Evening

Thorgott

Ted indeed gave the answer

Ted Shifrin

No, no, we're still using $[a,b]$, not $[a,a]$.

Edward Evans

@Alessandro gonna look them up lol

Alessandro Codenotti

Snarky Puppy - Lingus (We Like It Here)

►

Edward Evans

got it up on spotify hehe

Stupid Questions Inc

17:53

@TedShifrin i think i get it now thanks!

Ted Shifrin

No problem. Thanks to Thorgott too, I'm sure :)

@StupidQuestionsInc: That book is very difficult if you're not super experienced in mathematics proofs.

Stupid Questions Inc

@TedShifrin I know but I think i will be fine and it's worth it anyways :)

Edward Evans

@Alessandro very cool

Stupid Questions Inc

@TedShifrin btw do u still have contacts with Michael Spivak?

Ted Shifrin

Not in a few years.

Edward Evans

18:01

@Alessandro do you know Stimpy Lockjaw? This reminds me of them

Alessandro Codenotti

I don't know them, I plan to study a bit after dinner though and I need something to listen to

Ted Shifrin

They do get credit for great names :P

Edward Evans

Ah nice, I recommend Asteroids by them, veeeery cool

@TedShifrin I think Stimpy Lockjaw got their name from a random name generator lol

Ted Shifrin

LOL, aha.

Edward Evans

:D

modular forms and jazz fusion is a weird combo

Ted Shifrin

18:06

You just need a different congruence subgroup.

Edward Evans

i'm in level one at the moment :(

Thorgott

@StupidQuestions the same argument is given more explicitly here

sometimes a different exposition may help your understanding

I also like the accepted answer to that question

The techniques in both of these proofs appear pretty regularly in analysis

Edward Evans

18:27

@Lukas really not a big fan of all this inclusion rechnerei in the general Hecke theory stuff lol

I think it'll all make a lot more sense to me once I start hitting modular forms with Hecke operators

Semiclassical

18:50

@AkivaWeinberger one of the players eventually falls asleep and therefore the game ends.

(my guess for the resolution would be that the Master Game can't properly be regarded as a game at all, in the same way that one escapes Russell's paradox by insisting that the "set of all sets that aren't members of themselves" isn't actually a set)

(and looking at the description of hypergame, i see that's a rather obvious parallel)

Akiva Weinberger

19:14

@Semiclassical Ah, so that's what it's actually called

And you can make the parallel between them stronger by replacing "finite game" with the weaker condition "game that can never return to an identical state"

Semiclassical

nice

Akiva Weinberger

And then if the hypergame (with this changed definition) can never return to an identical state, then you can immediately ask to play the hypergame

Alternatively, you can make a parallel to the halting problem

(which isn't a paradox but you can do it anyway)

Semiclassical

nice

Akiva Weinberger

Restrict yourself to computable games: games where there exists an algorithm that decides whether a given move is legal or not

(which also implies each move has to be able to be represented in some finite alphabet)

If there is an algorithm that decides whether a given game is finite, then hypergame would be a computable game (and in fact we'd be able to write it down explicitly in any programming language)

But then the same logic as the paradox tells us hypergame can neither be finite nor infinite

We can't get out of this by saying "hypergame can't exist" because we can write it explicitly

Thus the only logical outcome is that there's no algorithm that decides if a given game is finite or not

and this implies that the halting problem is undecidable

Thus: Russell's paradox <=> hypergame paradox <=> halting problem undecidable

I think we can throw in a few more (Gödel's Incompleteness Theorems, uncountability of the reals / Cantor's theorem) into that set of equivalences

The Burali-Forti paradox as well (supremum of all ordinals is a larger ordinal‽)

Semiclassical

yeah, B-F is referenced in the paper's appendix

Akiva Weinberger

19:27

What paper

Oh the JSTOR article in the description of that link

Stupid Questions Inc

@Thorgott thanks i'm looking into it

Akiva Weinberger

You can probably connect these to the Barry paradox as well (Consider the smallest integer larger than all integers describable in fewer than a million letters)

Indeed, I think determining whether a given description is valid is equivalent to the halting problem

Semiclassical

@AkivaWeinberger yeah

it's in the appendix

Akiva Weinberger

20:00

Turing jump

In computability theory, the Turing jump or Turing jump operator, named for Alan Turing, is an operation that assigns to each decision problem X a successively harder decision problem X ′ with the property that X ′ is not decidable by an oracle machine with an oracle for X. The operator is called a jump operator because it increases the Turing degree of the problem X. That is, the problem X ′ is not Turing reducible to X. Post's theorem establishes a relationship between the Turing jump operator and the arithmetical hierarchy of sets of natural numbers. Informally, given a problem, the Turing jump...

Ooh

Shootforthemoon

20:37

Any suggestion (chat.stackexchange.com/transcript/message/53321561#53321561) for this? (with $a>0$*)

Thorgott

Cauchy condensation test or compare with an integral

Secret

21:27

So random question. If we plot the riemannian zeta on the surcomplex plane, will we find extra zeros due to the surrcomplex numbers that are now available to be plugged in?

Shootforthemoon

22:00

@Thorgott there's no way to use a comparison inequality with some telescoping series?

Thorgott

who knows

Shootforthemoon

ok, I'll go via condensation then

Semiclassical

23:16

@Secret take a look at this: mathoverflow.net/a/131816/55904

In particular: “ Nicolau claims that most standard functions, and even functions like the Riemann Zeta function should be obtainable in this manner.”

(See also the exercise in the linked sci-math post, lol)

That said, the post only suggests that one can get a surcomplex extension of Riemann zeta function. It doesn’t actually care it out. So I wouldn’t be surprised if no one has actually carried it out

(And I haven’t seen any other signs of such on Google)

Perturbative

Hey @TedShifrin :)

Ted Shifrin

Hi, @Perturb.

Perturbative

Do you mind if I ask for some general-ish grad school advice/career advice?

I just need to head out quickly but I'll be back in like 10 mins

Ted Shifrin

23:37

OK.

anakhro

Hi all.

Perturbative

@TedShifrin In the US do people who end up with professorships have to devote a lot of time to teaching? Is the split closer to 80/20 research to teaching or more like 60/40 research to teaching

anakhro

Does "logistics and planning" fall under teaching, or do you just mean lecture time?

Perturbative

Yeah I'd count logistics and planning

anakhro

Because lots of professors have various takes on being a teacher that either result in them doing a tonne of work, or very little besides the lectures.

And definitely depends on the course, if they have taught it before, etc.

If they have TAs for marking or what not.

Perturbative

23:54

Oh yeah definitely. For the sake of the argument, let's assume these professors are at research-focused institutions and have given the course a few times so their logistics and planning is at a minimal and there are TA's for marking.

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