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12:03 AM
From what I know, Thurston gave (many!) talks about the geometrization program and his proofs long before he wrote them up
And there was a collaborating effort from a wide community of mathematicians to follow through with the program, tie up the strands, simplify and develop the ideas, ...
At some point he decided that the community activity was a more enriching endeavor than just one guy writing up some stuff
12:53 AM
I just want to say that I adore Maths so damn much!
4 hours later…
5:01 AM
@BalarkaSen yes, but I meant there is(are) still some "big" open problem(s) on pseudo-Anosov theory on surfaces whereas most of the "big" open problems in (hyperbolic) 3-manifold are solved.
@BalarkaSen I once studied a bit but not now. Many (very) good researchers are already involved in this topic. But if you want to know the cornerstone paper of this theory, I should suggest Agol's paper: arxiv.org/abs/1008.1606
Veering triangulation is basically a taut ideal triangulation with an edge coloring (usually blue and red). Agol didn't define in this way but turned out two definitions are equivalent. Agol constructed a periodic splitting sequence of train tracks under the action of pA and used this to construct a triangulation of the mapping tori with pA monodromy.
it's a short paper you can read without much difficulty.
and as far as I know, there's a certain movement trying to give a "useful meaning" to the non-fibered faces of Thurston norm ball using "Veering polynomial" (this is an analog of the classical Teichmuller polynomial). If you want to know you can refer to Landry, Minsky, Taylor's Veering polynomial paper it's quite long btw.
@BalarkaSen what do you mean by "modern invariants"?
1 hour later…
6:29 AM
how to find iginvalue and iginvektor if matrisa nxn it same with 2x2 matrisa?if same find iginvalue and iginvektor from nxn wil be very hard?Is there an easier method?
2 hours later…
8:55 AM
Let a function $f:[a,b]\to\Bbb R$ be integrable on $[a,b].$ Then for each $x\in [a,b],$ $f$ is integrable on $[a,x],$ $\int_a^xf(t)dt$ exists and it depends on $x.$ Therefore, we can define a function $F$ on $[a,b]$ by $F(x)=\int_a^xf(t)dt.$
How do they know that $F$ is a function of $x$?
9:20 AM
Nvm, I got it.
9:34 AM
@onepotatotwopotato I think the big open problems in (hyp) 3-mfds are about codimension 1 structures on (hyp) 3-mfds now, not strictly about 3-mfds.
For instance: (pseudo)Anosov flows, taut foliations, tight contact structures.
@onepotatotwopotato Thanks!! Quite useful. I'll read the Agol paper
@onepotatotwopotato Heegaard-Floer, monopole Floer, instantons, pseudoholomorphic curves, ...
One would think that the non-fibered faces of the Thurston norm ball come from taut laminations in some way, yes?
If we are to believe the 2D story works in 3D, "limits" of incompressible surfaces should give rise to essential laminations. (But not conversely... many non-Haken hyperbolic 3-manifolds)
9:51 AM
@EE18 I think the statement is true.
@BalarkaSen I don't know much about non-fibered faces so no idea but it seems you're talking about Palark's taut polynomial?
We can translate the informal statement "there is no set that contains every other set implies the universe {π‘₯:π‘₯=π‘₯} is not a set" into a formal statement in ZFC like this:
¬βˆƒxβˆ€y(y∈x) β‡’ ¬βˆƒxβˆ€y((y=y)⇔(y∈x))
@BalarkaSen that's quite a satisfactory statement. Maybe such laminations are "dense" in some sense.
@BalarkaSen I know none of them but I know "Floer" thing is for knots. But good to know
2 hours later…
12:12 PM
The sum of five natural numbers, two of which are greater than 40, is 127. Accordingly, what is the largest of these numbers?

44, 52, 67, 76, 86.

Could it be that the correct answer is not even given? Since I get that

a+b+c+d+e = 127. a,b>40, this means that b=41 and c,d,e=1 so that a is maximized. This leaves us with a+41+3=127, so a=83 as the max value?
12:26 PM
I can calculate this limit using the 1^inf form but I don't get it intuitively. I mean cos(theta) can only take values less than or equal to 1 so if such a number is raised to infinity shouldn't the result always be zero.
12:38 PM
@Swan what do you mean 1^inf form? You can't calculate limits like that
Its should be zero but not by the reasons why you think its zero
@Jakobian I meant that if I consider that the limit is of the indeterminate form 1^infinity then I can rewrite the limit as e^((f(x)-1)(g(x)) and start simplifying further. But I think the value would be zero and it's not indeterminate at all.
@Jakobian What's the correct reason? Is it not that a number smaller than 1 raised to a huge number would tend to zero?
No, its more that, for $a>0$, we have that $(x/(1+x))^x\to e^{-1}$ so the whole limit is like $e^{-x}$
So the $x^2$ makes the powers way larger than the rate of convergence to $1$
Just because numbers are less than $1$ and powers go to infinity doesn't mean limit is zero. Isn't $(x/(1+x))^x$ an example of that
@Swan I don't like talking about indeterminate form at all. Its just a way of saying "you can't say anything about limit of this sequence from this and that alone". Very enlightening
1:05 PM
@THE_CRANIUM does your puzzle writer consider 0 natural?
@Jakobian note that this is just heuristic and you won't know until you calculate it
That definitely seems too few information.
1:27 PM
@Jakobian Does it now? I think that's balanced to be nontrivial, because for large x, cos(2 pi (x/(1+x))^a) is roughly (1 - (2 pi/a)^2/x^2), so raising it to power x^2 just balances it out to be some exp thing.
But I might be getting this wrong.
I think that ugly limit is written such that it's neither 0 nor 1 but makes the student work hard.
1:44 PM
Let V be a finitely-generated vector space and $T \in End(V)$. Given an arbitrary vector $v \in V$, we consider the minimal polynomial of $T$, $m_T$ and the cyclic subspace of $v$, $Z_v = Span(v, T(v), T^2(v), ...)$. Is it indeed true to say that the there exists a vector $u$ such that $Z_u$ spans a subspace of $V$ of dimension $deg(m_T)$ and it is the maximal dimension possible for a cyclic subspace? If yes, is this $u$ unique?
2:04 PM
@b_jonas Yeah, it's not zero. It's $e^{-2(Ο€a)^2}$. Thanks to both of you for the help 😊
2:34 PM
my heuristic was bad, I agree. In particular the approximation should be near $2\pi$ and not $0$
I wasn't really trying to focus on this particular example, rather to make a point
2:52 PM
ah yes, I think you're right and I messed up above. for large x, cos(2 pi (x/(1+x))^a) is roughly (1 - (2 pi a)^2/(2 x^2)).
1 hour later…
4:06 PM
A new agenda for low-dimensional topology, talking about the "Kirby's list".
4:20 PM
@onepotatotwopotato Are they going to tackle Meta Knight?
4:55 PM
@Thorgott idea on a positive answer to this question?
5:12 PM
quick question if possible
say i have elements a,b,c,d
is there a name for the specific type of partial order on the set of those elements which is such that $a \leq b \leq c \leq d$
is this a total order?
or, more strongly, a well ordering? I am familiar with total orders but not well orders
@Jakobian answer should be no
but im on way home rn
@Thorgott freakish answered it as 'no' (though without homeomorphism assumption). I think there should be conditions that would guarantee $p$ to be a covering map so the answer will be positive.
It's not my type of question so I can't add much. Feel free to mention me if you answer it
@EE18 its 0, 1, 2, 3
With Euclidean order
Yes its a well-order. It corresponds to ordinal $4$
You can call it order with order type 4
5:54 PM
perfect, thanks @Jakobian ! I don't know anything about ordinals yet so just the confirmation of it being a well order is enough for me, but looking forward to learning about ordinals later :)
Ordinals are order types of well-orders
That is, every well-order is isomorphic to a unique ordinal
They are 'representants' of well-orders
Just how cardinal numbers are representants of cardinalities of sets
6:12 PM
is it representants or representatives?
i think the latter but perhaps 'representants' is a technical word?
at any rate, thanks as always Jakobian
@Jakobian how would $p$ be a covering map? its fibers over the interior have cardinality 1 and that's not true over the boundary unless the identification is trivial
the reason i suspected it's false is cause every smooth manifold arises as quotient of a ball modulo some relations on the boundary
but i dont have an immediate counter-example and freakish's was already accepted, so i dont really care enough to dwell further on it
@Thorgott I never studied covering maps and so on, so that's why the stupidity on my part
6:31 PM
ah, no worries
"Numbers in mathematics are like time in physics: everyone knows what they are, and only experts find them hard to understand." Made me chuckle when I read it :)
@EE18 No, it's a French word.
6:46 PM
I didn't know my French had to be that sharp to keep up in this chat...will have to work on that...
You also need to know Chinese and a bit of German
Does anyone has any idea which theorem is referred to by "Intermediate Value theorem for Integrals" ?
Apparently a web search seems to be creating more confusions.
There are several, indeed.
Tons of theorems are popping up, which though interrelated, there should be a standard identofication of a theorem that goes by this name.
If $f$ is continuous on $[a,b]$, then $\int_a^b f(x)\,dx = f(c)(b-a)$ for some $c\in [a,b]$.
Who are you to say there should be a standard one?
6:56 PM
@TedShifrin a human being?
You're the supreme court of mathematical theorems?
Seriously, stop your whining.
@TedShifrin you're getting me wrong. I am not complaining and neither am I whining. I am just saying some of my thoughts because of the confusion I landed up in. If something's wrong with my thinking, why should I fear? There are always helpful people like you who will rectify me and guide me.
I am happy as long as I can learn from experienced and helpful people like you.
I suspect the set of theorems which are uniquely given a name is rather small
@TedShifrin Oh, so this is the standard interpretation, I presume. It seems then I am aware of this result/theorem. Thanks as always, @TedShifrin. I surely get it now.
@ThomasFinley just how there are lots of isomorphism theorems, there is lots of mean value theorems for integrals
7:03 PM
A question for those better versed in logic than I (and given that I do not intend to study logic for some time but would like to know if I am "cheating" when I do the following): when I write something like $\forall x P(x)$, am I being imprecise when I write the $\forall x$ part? That is, must I have previously defined some "domain" for the $x$s? By analogy (however crude) I think for example of ZF set theory wherein I must always specify from some other set.
@Jakobian Ok, this eliminates the confusion totally. You just gave a good analogy. Thank you!
Some of them concern continuous functions like in Ted's example, others are about $\int_a^b f(x)g(x)dx = g(a)\int_a^c f(x)dx + g(b)\int_c^b f(x)dx$, I think it was
This is all motivated by my trying to understand why the definition of notation like $\forall x \in X (P(x)) := \forall x ((x \in X) \land P(x))$ is rather useless (since it is necessarily false, not all $x$ are in any set $X$) and hence why the convention for that notation is $\forall x \in X (P(x)) := \forall x ((x \in X) \implies P(x))$
But my "understanding" above is only correct if writing something like $\forall x$ even makes sense
anyway, this is equally valid to interpret mean value theorem to mean that, and the one I mentioned is technically more important
in particular it has applications to Fourier theory
maybe for slightly different reasons, but both would be called mean value theorems for integrals by people
@Jakobian hmm...Thanks for all the references you gave! Indeed, they were helpful (,though I haven't studied Fourier theory as of yet).
7:10 PM
@EE18 $\forall x$ means quantification over all sets, and it is valid in set theory
got it. and this is a fact from mathematical logic?
i.e. it is allowed to speak/quantify over the "universe of all things"
(allowed in the sense that one does not run into any contradictions)
not sure what you are saying to me right now
all I said is that you can write $\forall x$
@EE18 its not only "useless" but not what we want
$\forall x ((x\in X\land P(x))$ is simply not what one would like to express when writing $\forall x\in X (P(x))$
@EE18 from perspective of language of ZFC there is no such thing as universe of all things
@Jakobian I'm not sure how I follow that $\forall x$ makes sense if there is no such universe then?
To be clear, I follow from ZF how there can be no set of all things
But I do not follow how we can say $\forall x$ but not be keeping in mind a universe of all things?
from the logical standpoint, $\forall x$ is just a quantifier
there is no such thing as a set
EE18 I thought your goal was to learn analysis. Why are you going down this rabbit hole?
7:22 PM
set is what we refer, informally, as objects of our study of set theory
It is and continues to be. So far the analysis is not giving me trouble though. It's Halmos which is giving me trouble (reading it alongside my analysis)
I should say, I am not yet at the analysis...in my analysis book(s) we are talking about real numbers at the moment
there is no such thing as universe of all sets, for there is no such object in ZFC
there is no such "set"
from perspective of ZFC, you are quantifying over all sets, not over some universe or a class
OK, that makes sense. To Ted's point, I don't want or need to go down the rabbit hole, but it's enough for me to know that writing $\forall x$ is not cheating. I will read Ebbinghaus or something like that one day
Thank you jakobian
From perspective of ZFC, you are allowed to write $\forall x$ in your language
you are allowed to conclude and manipulate logical expressions involving variables and symbol $\in$
you are allowed to write proofs using axioms of ZFC
the logic doesn't "think" of an universe, this is just an expression
you're quantifying over a variable, a variable which particular instance of we call a set. So you are quantifying over all sets
Got it, thank you!
Ted, to follow up, I guess I like having a precise understanding of the "interface" of what I am doing with the lower level stuff. Definitely agree that I don't want to spend too much time right now on said stuff, but at least for me I like having as good a sense as I can of that which I don't know (so that I can later make contact with what I do know). Hopefully that is a decent approach
7:32 PM
If you want to know my opinion, I think it will be useless for you
what's the "it" there?
no one should study set theory unless they're interested in it, or it appears in what they're interested about
I'm just studying the naive version though :)
Let's put it this way, EE18. I've been a teaching and research mathematician for 50+ years. I've taught undergraduates the basics of foundational mathematics so that they can study mathematics. I took only half of one course in mathematical logic and then dropped it. There are indeed lots of people who are obsessed with these foundational issues, but most mathematicians know how to use quantifiers without needing all these rabbit holes.
@EE18 you seem to be asking a lot of questions about the non-naive version of set theory for someone studying naive set theory
7:36 PM
theres nothing naive about naive set theory tbh
That's well taken Ted. I will continue to fight the urge to get sucked into said holes :)
I use naive just because Halmos does, but perhaps Thorgott when you say naive set theory you're referring to the book itself?
1 hour later…
8:39 PM

 Set theory

Anything related to set theory. For instructions how to render...
9:28 PM
I graduated today.
@DannyuNDos Congratulations!!
Congratulations πŸŽ‰πŸ‘πŸ»πŸ‘¨πŸ»β€πŸŽ“
whats your paper about
Complete metric spaces.
9:37 PM
By the way, "into the complete metric spaces" looks a bit weird
maybe that's just me
@DannyuNDos yeah but what about them. Of course, I can read the title myself
@DannyuNDos Congrats!
@Jakobian Tbh, it's just a survey of exercises which I proposed solutions to.
Yes, the β€œthe” is not proper English.
Probably too late to remove the "the", eh?
My advisor suggested the "the" so whatever
9:41 PM
@DannyuNDos I see. From Munkres or something?
The "the" sounds context driven
This is what happens when non-native speakers cannot do definite or indefinite articles.
Asians and Russians have no clue.
I'm proud of myself for catching this error (I did need confirmation from Ted, though)
Reading it will provide the justification
9:44 PM
My intuition is right at least
about English
On the surface, yes
there is no depth to it though
not as far as I can tell
@DannyuNDos are you planning to study spaces of isometries? Or its just a one time thing
Your English, like @Thor’s, @Astyx’s, @AlessandroCodenotti,s, and that of many others, is excellent.
@Jakobian It is a one time thing, I think.
While writing this, I was gonna cover the entire Chapter 7 of Munkres, but my advisor suggested that covering one section is enough for a master's thesis.
My Bachelor thesis was about fractional directional derivatives, and their integral representation using a kernel
For $\alpha$-harmonic functions
I've proved the formula for them
They're applied to some boundary problems... like the Laplace equation I think, but fractional, with moving boundary
A lot of people might be surprised it was about analysis and not something like general topology
10:03 PM
Hi! $f:\mathbb R^n\to\mathbb R^n$ continuous bijective. How to prove that if $f(K)$ is compact then $K$ is compact.
@PNDas this is true for any continuous function
If $f:X\to Y$ is continuous, $K\subseteq X$ compact, then $f(K)$ is compact
@Jakobian No.
oh the other way
@Jakobian I wrote the opposite.
If $f$ is continuous and bijective from $\mathbb{R}^n$ to itself, then $f$ is a homeomorphism
from open mapping theorem
10:04 PM
@Jakobian Actually, I am using that to prove this. math.stackexchange.com/a/59537/612798
$K$ must be closed, prove its bounded
that probably works
I would now like to classify this as a reasonable idea which doesn't seem to pan out. Sorry for giving it as "hints". But there are actually some interesting issues here which I hope to return to eventually... — Pete L. Clark Aug 25, 2011 at 16:18
author claims he made a mistake
It makes sense. This would lead to a simple proof of open mapping theorem which doesn't seem to exist
Actually I want a proof using Invariance of domain. "The argument I originally had in mind turned out to need Invariance of Domain"-Pete.
10:20 PM
Sorry to bring back an oldie (but goodie). In preparation for reading about the inversion formula for the Fourier transform, I'm going over the integral $\int_{0}^{\infty}\frac{\sin x}{x}dx$. The method I'd like to understand is in evaluating this through double integration, i.e. $$\lim_{s \to \infty}\int_{0}^{s}\frac{\sin x}{x}dx= \lim_{s \to \infty}\int_{0}^{s}\int_{0}^{\infty}e^{-xt}\sin x \ dtdx.$$
And then suddenly...out of the blue...cold turkey...the order of integration is switched. I have not found a decent explanation of this. If this is super clear to someone, I'd be grateful to hear from you.
@PNDas By invariance of domain $f$ is a homeomorphism. This makes it trivial
Invariance of domain is what I call open mapping theorem above
@TedShifrin not sure if you meant me, but thank you if you did
I appreciate the complement
@psie you can justify it using Fubini theorem
suppose you fix $s$, then show $e^{-xt}\sin(x)$ is absolutely integrable
@Jakobian you mean on $[0,s)\times [0,\infty)$?
actually I don't know
@psie where did you find it anyway?
In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet, one of which is the improper integral of the sinc function over the positive real line: This integral is not absolutely convergent, meaning | sin ⁑ x x | {\displaystyle \left|{\frac {\sin x}{x}}\right|} has infinite Lebesgue or Riemann improper integral...
they claim the integral is absolutely convergent, it seems
10:35 PM
@Jakobian Of course I did … in this instance. :)
It's also cool that the square has integral pi/2. math.stackexchange.com/q/13344/612798
@psie I don't trust that section on wikipedia. It has weird notation
@psie yeah, it's confusing, because in the top they say it is not absolutely convergent, but then in the subsection "...for all $s>0$, the integral is absolutely convergent."
@Jakobian In retribution, it’s compliment ;)
nah that was fine too. if you're too good at spelling in English, you're a spy
10:40 PM
Anyway, I believe double integration can still be applied, the change of order of integration is just a very subtle argument it seems
psie any amount of time you could spend on reading non wikipedia sources about the Fourier transform would be amply repaid in understanding
keep in mind that 99% of people who fourier transform stuff are engineers in some class where the professor is a formula book
i'm not even convinced that they use it in real life
I also put random "s" at the end of sentences, words disappear from what I write, and I replace words with different words, like thing and think getting replaced just because I think they sound somewhat similar.
ok, boycott Wikipedia
@psie In many instances, that is warranted. Books exist for a reason.
wikipedia is just going to be predictably not-math-oriented and maybe not 100% math-reliable in math topics that are mainly used by people who don't do math
see also a ton of its coverage of linear algebra
i'm sure its coverage of category theory is fine, as far as all of that goes, because the 5 people who edit it full time are probably also category theorists
10:44 PM
There have been so many questions lately on this site with people thinking that the kernel and image of a linear map are subspaces of the same vector space. What's with this?
not so, with a subject that literally every engineer meets several times in college
ted that is odd i ran into that the other day
11:00 PM
I have encountered it several times of late.
Now I just encountered someone posting his smattering of solutions to textbook problems as solutions manuals. I'd noticed google giving me solutions manuals to my own textbooks earlier. There are so many people out there bilking innocent math students.
What's the word for psychological occurrence of consistently replacing words with similar sounding ones?
that's odd enough of a thing that i don't know such a term
ted: i will continue to sell my solutions manuals and there's nothing you can do to stop it
A specific occurrence of this seems to be malapropism. But I swear I saw the name for it somewhere
You can sue you for me, leslie.
Leslie suing himself?
11:14 PM
Why not?
in Athens, GA, there is a tree that owns itself.
the insurance market takes care of a good portion of what we would want to achieve by suing ourselves, although you do occasionally still see people suing their friends or family members to trigger coverage under a policy
i think bar rules would prevent me from representing myself in a suit against myself because of my conflict of interest, so i would have to proceed pro se
Q: Why do so many solutions to $n+1\mid3^n+1$ satisfy $n\equiv27\pmod{72}$?

Akiva WeinbergerHere are the first $40$ integers for which $\dfrac{3^n+1}{n+1}$ is an integer. I've denoted their residue mod $72$ by color and a symbol. $$\begin{array}{r}\dagger\,\color{violet}0,&\bullet\,\color{brown}1,&*\,\color{red}3,&27,&531,\\ 1\,035,&4\,635,&6\,363,&11\,475,&19\,683,\\ 4\,131,&80\,955,&*...

@Jakobian I don't know about the psychological condition, but the relation between said words is I believe referred to as homonym(s)
I have never understood why one can only be represented by a member of the bar
why is one not allowed to waive that?
EE18: i dunno about in elsewhere, but in the US, you are?
even in this country where so much is about not having the state be too paternalistic
oh are you?
I was under the impression that only members of the bar could be used as counsel?
perhaps i should have done a cursory google search though
11:19 PM
if you're a human being (and not a business entity) you can represent yourself
you can't get your cousin who is not a lawyer to represent you in court
that doesn't mean your cousin can't advise your whole legal strategy and tell you what to say, he just can't question witnesses, talk to the judge, etc
that's what I mean, sorry. why not? what's the reason against this (if someone is of sound mind and makes that decision)?
the simplest answer to "why not" is just "because it's illegal," but there are various policy rationales, including that your cousin may not do a very good job of representing your interests if he is not trained in the relevant law, talking to judge, etc.
which is paternalistic but so are most licensing regimes
more importantly, your cousin might abuse witnesses and otherwise waste a court's time if he were allowed to do what counsel get to do (sometimes have to do)
ya i guess I've always wondered about the policy rationales :) not even being inclined to such things, i had always assumed there was some conspiring on the part of the bar to arrange this
note also, that's allowed for people representing themselves, and they do often abuse witnesses and waste the court's time
so it kind of protects the public to only permit parties to get to do that, and not literally anybody with some desire to do that in court
does being a member of the bar preclude such things anymore than simply banning anyone found to be in contempt from ever representing someone else again?
any more than*
11:25 PM
well, it gives those people more to lose
it doesn't stop humans from being bad any more than any law stops anybody from doing anything
but on average, even a malevolent/incompetent lawyer is less abusive than a random member of the public who wants to tangle in court probably would be, if they were allowed to do so
that's well taken
a new way to answer that question for me now!
at least, i used to volunteer pretty regularly at a clinic for people who were representing themselves in civil litigation, and they tended to have the worst ideas
i guess it's just a surprising state of affairs for a country where, in many respects, the MO is "the individual can do more or less what they want; even when that liberty comes at great cost, it is worth it"
Would I be thrown out for contemptible in court?
certainly if you came after me and asked the court to second-guess the honest American pastime of selling solutions manuals
11:37 PM
Anyone here have an OEIS account?
Nigel Richards is something else
akiva, i was just about to ask if you'd posted more numerical evidence somewhere (i realize MSE isn't really built for this)
@AkivaWeinberger yes
howdy @ DogAteMy
11:42 PM
Swedish = Mats
@leslietownes Do you mean you're suing yourself for such a stupid reason that you can't find an ordinary lawyer who'll represent you?
haha, for the record, ted was the one who brought up the possibility of me suing myself on his behalf. i'm not suing anybody
Funny how people take everything out of context ...
ted would like to see me punished for selling sub-standard solutions manuals to his textbooks (in this fictional narrative), and rather than trouble the district attorney with it, he'd rather i sue myself
@Jakobian Does this mean you do?
11:46 PM
@AkivaWeinberger no. It means Mats does
I don't have one

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