12:12 AM
Are you asking for a concrete procedure?

12:37 AM
An algorithm

1:18 AM
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?

1:29 AM
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.

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

2:27 AM
Is there a way I can delete a question of mine here?
After a while just shows flag for moderator

3:26 AM
Unless it is profane, I don't see why you'd want to delete it.

4:11 AM
en.wikipedia.org/wiki/Graphon generalizes immediately to directed graphs if the graphon is asymmetric, right?

4:29 AM
Feb 10 '17 at 21:45, by Akiva Weinberger
Graph theory game http://treksit.com/?thegame

5:12 AM
@AkivaWeinberger so basically planarity.net but with background effects

5:30 AM
also, the bonus area is creepy af

6:06 AM
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?

6:29 AM
@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?

6:52 AM
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.

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?

7:09 AM
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

7:40 AM
@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.

Oh, that's meta

2 hours later…
10:14 AM
@MaryStar you should rethink the "$|\frac{x}{x}-1|=|\frac{x-1}{x}|$"

10:45 AM
0

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... 11:28 AM 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. $Any suggestion? 11:46 AM @EdwardEvans suggest me some cool metal 12:02 PM @AlessandroCodenotti what kind of Metal? Hahaha Or rather, do you most often listen to metal? So I know not to recommend some obvious bands lol I'd say I listen mostly to prog rock and post rock, but I also listen to a lot of metal @EdwardEvans Something heavy but technical maybe Something that Balarka would like let's say lol Lately I've been getting into Cult of Luna for example 12:22 PM Lol fair, I listen mostly to black metal, but do you know The Zenith Passage? @AlessandroCodenotti the Song Deus Deceptor is cool af @EdwardEvans no, I'll check them out, thanks! I'm not very familiar with black metal but if you have cool stuff to suggest I'm always looking forward to listen to something new Nice one, also Fallujah are great, check out Sapphire or Chemical cave from that album I will, thanks! And if you want some „post black metal“ which is more accessible then Numenorean are incredible, and møl Hehe I know Numenorean, I listen to the last album when it came out and liked it Didn't expect to see them mentioned here! 12:40 PM Yeah Numenorean are great, they played in Mainz recently Nice There's a lot of interesting concerts in Köln, but they always happen while I'm in Italy somehow There's a group from NRW that I really like, I missed them last year when they played in Bochum and I'll miss them again in April when they'll play in Köln, both times because I'm in Italy But I'll go to a Snarky Puppy concert while in Italy in April so that's fine :P Nice, I’m seeing Helrunar tomorrow Loka Lögsaga is Great But black black metal lol 2 hours later… 2:32 PM Yo @Lukas, I have a "general" Hecke theory question for ya if you're here :P okay So suppose$(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 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 Ahhhhhhhh so I'm thinking of$R\setminus S / R$as "$(R\setminus S)/R$" lol yes exactly that 2:41 PM omfg hahaha this gives that every left coset is contained in a unique double coset okay makes sense, thanks :P and so you get that equality thanks man :) 3:40 PM 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$U$" 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$n$that gets it done, thanks @AlessandroCodenotti why is that like the standard example for compactness 3:57 PM I guess because it's the easiest example after a closed interval 4:12 PM 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. 4:50 PM 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) What would happen if there were no such$x$? 5:09 PM @Thorgott then$\sup A = a\alpha=\sup A$by definition in either case What are the properties uniquely defining$\alpha$? 5:37 PM @Thorgott if$x\in A$then$x\leq \alpha$but how do we know that there's some$x\in A$and$x\neq a$@Alessandro: What about Snarky Kitty? 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 @StupidQuestions that's the definition of an upper bound, but the supremum is a more specific one among these I figured there should be a snarky kitty to go with it, @Alessandro :P @Thorgott yes, if$b$is an upper bound to$A$then$\alpha\leq b$5:43 PM @StupidQuestionsInc: I haven't been following your discussion, so forgive the interruption. But if$A=\{a\}$, then of course$\sup A = a$and there's nobody smaller in$A$. @TedShifrin yes, the non-trivial part that I'm not getting is why there should be an$x\in A\setminus\{a\}$in the first place @StupidQuestion right, so if no$x$in an interval to the left of$\alpha$is in$A$, then..$\alpha=\sup A =a$No, how did you get that? You're talking about Spivak's proof of Theorem 1-3? 5:45 PM well since there's no$x$to the left of$\sup A$then that forces$A$to be as small as possible, i.e.$A=\{a\}$@TedShifrin exactly Well, look at the definition of the set$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$(why?). No, we're not talking about all possible$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$, then.. @TedShifrin that's exactly the part I'm missing I don't see why$\alpha > a$Because of the definition of the set$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$, since you didn't include it in your picture.) Hi @Mats Hi Ted. 5:49 PM @Thorgott but in the definition of$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$(apologies for not including the picture of the preceding page) I just told you that! @TedShifrin let me take a rest and then come back to it maybe I will get an aha moment then It's an open cover, so any open set (which is made up of open intervals, by the way) containing$a$must contain a whole interval to its right. @TedShifrin oooh, so the fact that$[a,a]=\{a\}$has a finite cover means that there's some open interval containing it That's different from the question you initially posted 5:51 PM Evening Ted indeed gave the answer No, no, we're still using$[a,b]$, not$[a,a]$. @Alessandro gonna look them up lol got it up on spotify hehe 5:53 PM @TedShifrin i think i get it now thanks! No problem. Thanks to Thorgott too, I'm sure :) @StupidQuestionsInc: That book is very difficult if you're not super experienced in mathematics proofs. @TedShifrin I know but I think i will be fine and it's worth it anyways :) @Alessandro very cool @TedShifrin btw do u still have contacts with Michael Spivak? Not in a few years. 6:01 PM @Alessandro do you know Stimpy Lockjaw? This reminds me of them I don't know them, I plan to study a bit after dinner though and I need something to listen to They do get credit for great names :P Ah nice, I recommend Asteroids by them, veeeery cool @TedShifrin I think Stimpy Lockjaw got their name from a random name generator lol LOL, aha. :D modular forms and jazz fusion is a weird combo 6:06 PM You just need a different congruence subgroup. i'm in level one at the moment :( @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 6:27 PM @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 6:50 PM @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) 7:14 PM @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" nice 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) nice 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‽) yeah, B-F is referenced in the paper's appendix 7:27 PM What paper Oh the JSTOR article in the description of that link @Thorgott thanks i'm looking into it 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 @AkivaWeinberger yeah it's in the appendix 8:00 PM 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 8:37 PM Any suggestion (chat.stackexchange.com/transcript/message/53321561#53321561) for this? (with$a>0\$*)

Cauchy condensation test or compare with an integral

9:27 PM
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?

10:00 PM
@Thorgott there's no way to use a comparison inequality with some telescoping series?

who knows

ok, I'll go via condensation then

1 hour later…
11:16 PM
@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.”
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)

Hey @TedShifrin :)

Hi, @Perturb.

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

11:37 PM
OK.

Hi all.

@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

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

Yeah I'd count logistics and planning

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.

11:54 PM
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.