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10:07
Tout le monde parle français ici en fait
@LeGrandDODOM si, les autres seulement faisent semblant qu'ils ne parlent pas francais
@Vrouvrou if you want an interesting question related to this topics: it's easy to show that in a $T_2$ space if a sequence has a limit it is unique. What about the converse, if a space is $T_1$ and all the converging sequences have a unique limit must it be $T_2$?
@mercio, is this correct
Let $f$ be continuous on a domain (open and connected), $D$. Now suppose that for every $z_0 \in D$, there exists $B_\delta (z_0)$ such that $f(z)$ is constant in $B_\delta (z_0)$. Show that $f$ is constant in $D$.

I don't know if this is correct, but allow me to share my thoughts. If $f$ is continuous on $D$, then $f$ is continuous on every $z_0 \in D$. Define an open ball, $B_\delta (z_0)$, such that $f(z) = K$, for all $z \in B_\delta (z_0)$.

Since $D$ is a region, it is connected, then $D\ne S_1 \cup S_2$, where $S_1 \cap S_2 = \emptyset$. Then, $D$ is the un
K ?
I don't think it makes sense
@AlessandroCodenotti che sono $T_1$ e $T_2$?
10:18
$T_2$ (or Hausdorff) means every pair of points has a pair of open disjoint neighbourhoods, while $T_1$ means that for every pair of points there is neighbourhood of the first not containing the second and viceversa (equivalently a space is $T_1$ if the singletons are closed sets)
10:34
@DHMO Try these clogs for size: math.stackexchange.com/questions/2117616/…
@RobertFrost could you give me an example?
@mercio, is there a proof not involving much topology? math.stackexchange.com/questions/2117596/…
the concepts themselves involve topology so idk what you're expecting
@mercio, how come the requirement for the function to be continuous is not used in the proof? Does this mean that continuity is not required?
continuity is not required
also it's very implied by the "locally constant" hypothesis
so it's not that it's not required, rather it is redundant
locally constant is a lot stronger than merely continuous
10:46
I have a faint understanding of the proof given by Andre. But I don't understand why the path between the two points, x and y given by gamma is a function from [0,1] to D. From there, I understand that the image of gamma is compact, since [0,1] itself is closed and bounded, and the continuous image of a compact set is compact.
By the way, thanks so much @mercio! You're such a great guy.
that's the definition of a path @Harry
in general, connected doesn't imply path-connected though
the simplest proof really is to prove that $f^{-1}(\{y\})$ is open and closed for every $y$
and if $D$ is not empty then one of them is non empty
which means that one is equal to $D$
because $D$ is connected
Evaluate $\displaystyle\int\frac{x^2+x}{(e^x+x+1)^2}\ \mathrm dx$ (no cheating!)
I find it weird that the chatroom does not have MathJax hahaha
it has
tinyurl.com/cfqcvpc
10:50
@AlessandroCodenotti, as I said, I don't have much topology hahaha
@mercio, I see the code in @DHMO's post, though
Hey everyone!
@DHMO an example of the question?
@RobertFrost yes
an example of $x$ that satisfies the property
Take the number x_0=10, 2 factors this once. So x_1=3*10+2=32. o_1=1 is the product of the odd factors of x_1. 1 is coprime with 5, as per the proposition
I see
10:56
I don't
In the uniform topology on $\mathbb{R}^{\omega}$, $\rho (x, y)$ is always at most $1$ for any $x, y \in \mathbb{R}^{\omega}$ correct?
why did you put $3*10+2^1$ and not $3*10+2^5$ ?
@mercio 2^1 is the highest factor of 10
but in his post he says $2^{o_n}$
and if $x_n=10$ then $o_n=5$
10:59
When we have the transformation matrix of $\Phi$ in relation to $B$, when to calculate the transformation matrix of $\Phi\circ\Phi$ in relation to $B$, we have to calculate the images $\Phi^2(b_i), b_i\in B$, right?
and no @RobertFrost you are not going to prove that there are no nontrivial collatz cycles like this
@mercio The truth or falsehood of that statement lies in the answer to my question.
yes I'm only telling you in advance
@mercio you know little about me.
@mercio i'm special ;)
dang :o
orz
@RobertFrost if I follow this correctly, you expect that the odd integers in collatz sequences are all pairwise coprime ?
that seems highly unlikely to me
if you start on 1000000000000000000005, I would be very surprised if you don't get to another multiple of 5
x_0=m.2^n
x_1=(3m+1)2^n
make your case that m and 3m+1 are coprime
@RobertFrost @mercio ^
what do you guys think about math.stackexchange.com/questions/2116141/…
seems overwhelmingly unlikely that it loops back to $0$
also lol i gave the worst try at a counter example with my 1000000000000000005
because at the start it will only mimic what happens to 5
11:26
@mercio indeed there is solution to x^2+1 = p
confirmed by using Legendre symbol
yeah
but whether the sequence can reach x is another problem
@Perturbative what's the uniform topology and what's $\rho$?
Thanks, @Perturbative!
Hey guys is the following true (We only speak about real matrices):
1.) If A is symmetric all eigenvalues are real.
2.) If B is a real matrix and C = B*B^T & C' = B^T*B => Eigenvalues of C&C' are non-negative.
3.) 1.)&2.) => There are symmetric matrices that can't be written as A = B*B^T
4.) and those matrices have negatives eigenvalues? Or isn't it possible to say something about their eigenvalues?
11:41
@mercio Is that a counterexample?
well it should
just like almost any odd large multiple of 5
1) is true, 2) sounds true, 3) is true, 4) sounds true
(assuming all your matrices are over $\Bbb R$)
I'm not sure about 2 and 4, I'd need to think about them but it's lunchtime now
@AlessandroCodenotti I'd appreciate it but have a nice lunch first :)
Hello,I want to discuss pair of equations topic
Let L1=0 is one straight line equation and L2=0 another straight line equation then (L1)(L2)=0 represents pair of equations right?
Then L3 be another straight line equation,then after homogenising (L1)(L2) equation with L3 gives?
@Aresloom Perhaps we could move to linear algebra chat room so that we do not discuss several topic in the same room at the same time.
11:53
@MartinSleziak sure didn't know there exists a lin alg chat
@DHMO Yes I said they were equivalent. The formulation in the question is just a different way of factoring out the 2^n factors. I am claiming 3x+1 is coprime with x (obviously) and that 9x+3+1 is too if you disallow 2 as a factor and so on. I think it's a nice simplification.
12:24
@AlessandroCodenotti The definition for the uniform topology and the uniform metric is given above
ok so $\rho$ is actually $||\cdot||_\infty$ on $\Bbb R^\Bbb R$
no wait, that might be wrong, what's the standard bounded metric on $\Bbb R$?
@AlessandroCodenotti Out of curiosity, are the terms (e.g 'uniform metric') I'm using non-standard? Because the author of the text I'm learning from seems to use non-standard terminology quite often..
I don't know, I've never heard them before, but I've never seen another name for those things either
Which book is it?
12:43
Topology: A First Course by Munkres
Ok so it does look like $\rho(x,y)$ is at most $1$
I never used Munkres myself but I've heard a lot of people recommending it
@AlessandroCodenotti Which book(s) did you learn General Topology from?
I only chose Munkres because a lot of people recommended it
A book which is only available in Italian and I wouldn't really reccomend anyway, but I learnt most of what I know in topology from lectures, I had a very good professor
Does this bounded metric induce the same topology as the original metric?
Yep it does
It looked like it, that makes much more sense now
12:56
It would be fun to learn Topology from a book in Italian :p, would give me a good excuse to learn some Italian myself
Where are you from if you don't mind me asking?
I'm from South Africa
Wow, nice, that's further than I would have guessed!
13:12
@Perturbative Do you speak Afrikaans?
@Krijn Ek kan nie Afrikaans praat nie
Ah jammer, Nederlands en Afrikaans is vrij leesbaar voor elkaar
Ek dink German is eintlik soos Nederlands
@Krijn Just joking, I can speak (a little bit of) Afrikaans, it was one of the languages we had to learn in school (along with Zulu, which is also fun)
Lol
Really interesting language, Afrikaans
So weird for Dutchmen
@Krijn kann man, dass Afrikaans spreche, Deutsch verstanden?
13:24
verstehen @DHMO infinitiv after a modalverb
@AlessandroCodenotti oops
no so che parlai aleman
There's few Pure Mathematicians in South Africa though, sadly. To give an example there's only 2 known people doing research in Algebraic Geometry in the whole country
@DHMO I don't think so, Afrikaans is just funny sounding Dutch, German is a totally different language. But @Perturbative has a better answer probably?
@Krijn they're all related
Dutch and German are related
Of course
But Afrikaans and Dutch much more closely
13:28
yes that is true
Dutch is German with vowels
what the hell?
what's new
@Perturbative I can see eintlik = einlich right
Hi @Balarka, welcome back to the linguistics chat room
13:30
@BalarkaSen namaste
I prefer the art and interpretation chat room
@Krijn In Afrikaans objects are often referred to in masculine form for some unknown reason, does the same thing occur in Dutch? Like in English we would say "That's my boat, she's a beauty", analogously in Afrikaans one would say "That's my boat, he's a beauty"
@Perturbative Hmm, I do this as well, but there are female objects such as ships and corporations. So there are rules for this, but I tend to ignore them.
@Perturbative Afrikaans has no gender?
@DHMO Afrikaans is weird at times
13:40
@Perturbative maybe you should speak in Afrikaans and I speak in German (I only know a little German) and we see how much we understood each other
@DHMO Perhaps another time, a bit busy now :)
sure
So you know only from context whether you speak about a women or men? In afrikaans?
@Aresloom no, it just means that we refer to a boat as a male
just like Englishmen refer to countries as females
13:44
Afrikaans has hy and sy for he and she
and guess what "dit" means
"it" of course
here's the catch:
"his boat" is "sy boot"
so "sy" can mean "she" and "his" and "its"
ah I see what you mean :)
@DHMO "Hulle" is the same as "Zij" in Dutch or "They" in English; the funny thing is that it's also the same as "Hullie" in the regional dialect where I live
@Krijn I see
@Krijn why are you required to learn Afrikaans?
13:50
Oh I am not at all, I barely know any Afrikaans
It's just very close to Dutch and especially my dialect so I like reading it
In Italian possesive articles agree with the gender of the owned object rather than the owner (il suo libro could be either his book or her book, suo is male only because libro is male), in English they agree with the owner (which makes sense, nouns are genderless) and in German they agree with both (for the third person, just with the object for the others), how does that work in Dutch?
Reading wikipedia in Afrikaans is just brilliant, and on the Talk page sometimes Dutch people and Afrikaners have a conversation using just their own language
@AlessandroCodenotti in English they agree with the owner even before the genders are done away with
@AlessandroCodenotti As in English
So it's his car because the owner is male
@AlessandroCodenotti por que puedes hablar aleman?
13:55
zijn auto instead of haar auto
I lived for a year in Germany
and before you said, I never noticed that in German they agree with both... it's really crazy
@DHMO wieso redest du spanisch mit einem italiener?
Well I lived for a year in China too yet I forgot all the little Chinese I knew, I also frequented intesive courses in Germany
@s.harp er hat gesagt, dass er spanisch versteht
13:57
@s.harp I can read and understand most Spanish, it's very similar to Italian
(obwohl er spricht kein spanisch)
(solo queria hacer una broma)
wait, @AlessandroCodenotti how do they agree with both?
Word order, the verb goes at the end after obwohl
Relevant: What does "Uvuvwevwevwe Onyetenyevwe Ugwemubwem Ossas" mean in Afrikaan?
13:59
right
@BalarkaSen it's probably not Afrikaans
Africa has a lot of languages other than Afrikaans
and I can assure you that it is not Afrikaans
@Balarka the first word probably means "I hit my toe and it hurts" :)
Since about 30% of English comes from Latin and Latin doesn't have articles
it's always vaguely amused me that the determiner can actually reverse the meaning in cases like this. So "this method has a few flaws" kind of has the opposite connotation to "this method has few flaws".
Sometimes they agree with both, for example in akkusativ sein is for neutral objects with a male owner, ihren is for male objects with a female owner and so on
@s.harp is that out of a textbook?
14:08
No its some notes by gromov called "Spaces and questions"
Interesting.
Gromov's stuff are always interesting
its like hatcher on steroids: proof by mentioning a theorem that can be used in the proof
I don't really understand what you mean by that. (Hatcher is incidentally my favorite expositor.)
Gromov is not a good expositor, but a very good mathematician.
The way I read in Hatcher has always been very much talking and not so much care about explicitly seeing how a proof works. Yesterday I wanted to read up Hurewicz theorem and i went to the index of Hatcher, went to the relevant page and he does everything in reference to what he did before
Somethign along the lines of "We do the construction we did in this case and then apply it here and it works"
ie the author assumes the reader is on board with what he is thinking so does not bring a lot of detail in having a self contained exposition
when I read this and the other notes by gromov I get the same feeling but magnified by a lot: He has some thoughts or feelings about something and assumes that if he vaguely describes it the reader will know what he means
14:23
Hi chat
for example the question I have linked what looks to be a proof via a construction that is not standard is just abbreviated via "it follows from poincare bendixon"
Hatcher does prove everything very explicitly so I do not understand your complaint (which happens to be a popular one, too). Also, Hurewicz theorem literally is just a commutative diagram argument with the long exact sequences of homotopy and homology, which he talked about before.
Hatcher does not write conventionally, yes - he follows an intuition-based exposition. This is one of the two techniques in any of the standard algebraic topology textbooks you'll see, the other being category-theoretic which, I find, is sometimes unhelpful.
I'm not sure how to see if $\phi$ is a $1$-form on $\Bbb R^2$ so that $\lim_{\|x\|\to\infty}\phi \equiv dx_2$ that Poincare Bendixon implies the existence of a nowhere vanishing function $\rho$ that is asymptotically $1$ so that $d(\rho \phi)=0$
(that with regard to the gromov notes)
I haven't read the paragraph you are referring to, but I do believe you that Gromov is rather hard to follow. Even experts in his field(s) agree on that
It is possible that my complaint about hatcher does him injustice, and I haven't attempted to read his writing for ages except for looking up hurewicz the other day where I was frustrated by the lack of a simple "this is it, this is how its done"
14:33
If you know the homotopy and homology long exact sequences (which is what he refers back to essentially) it shouldn't be complicated to follow
My algebraic topology is extremly rusty, and after about 10 seconds I google searched for another source with which I was very satisfied
hi, is anyone here familiar with the conditional independence test algorithm? for example given a few conditional independence statements, how to construct the conditional independence graph.
tbh the exact oppositive of gromov is probably baez
??!!
maybe I see what you mean
14:52
Solve $ae^{bx} + cx = 0$ in terms of the Lambert W function.
15:11
In mathematics, the Lambert-W function, also called the omega function or product logarithm, is a set of functions, namely the branches of the inverse relation of the function f(z) = zez where ez is the exponential function and z is any complex number. In other words z = f − 1 ( z e z ) = W ( z e z ) ...
@DHMO Have seen the anti derivative of the Lambert W function?
yes
ok
@DHMO Do you think your good at evaluating a series?
depends on what series it is
Hm...
Have you tried calculating pi by hand before?
i don't think so
i know there's a quadratic precision algorithm out there
15:22
@DHMO Try the following integral:$$\int_0^1\frac1{1+x^2}\ dx$$
point P is on the terminal side of angle . Evaluate
the six trigonometric functions for . If the function is undefined,
write “undefined.”
@SimplyBeautifulArt that's slow
THE GIVEN P POINT IS (0,5)
@DHMO Then make it faster :)
I have having a hard time doing this
15:24
@SimplyBeautifulArt no idea how
Oh, come on
So can you solve the integral analytically?
it's pi/4 of course
and then taylor seris
i know this algorithm
@DHMO Well, you should take geometric series instead
it's the same
Mhm
And then you say the series converges to slowly
$$\frac\pi4=1-\frac13+\frac15-\frac17+\dots$$
15:28
ya
@DHMO Surely you know how to converge faster
@SimplyBeautifulArt Euler's transform!
Lol, you are "slow", but I can't accelerate your convergence
let's see
the best acceleration technique is cohen-villegas-zagier
15:30
1/(n-1)^2 - 1/(n+1)^2 = 4n/(n-1)^2(n+1)^2 right
Sure?
Not entirely sure what your doing
@BalarkaSen Interesting
Nah. The best series acceleration is to replace $a_0,a_0+a_1,...$ with $\sum_{n=0}^\infty a_n,0,0,0...$ :P
@BalarkaSen Sadly, it looks very complicated
@SimplyBeautifulArt what is your solution?
@DHMO Huh?
$$\sum_{k=0}^\infty\frac{(-1)^n}{2k+1}=\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0‌​}^n\binom nk\frac{(-1)^n}{2k+1}$$
15:33
@Simply Fair enough. It's easy to implement it
:-/ And then plugging in an upper bound on the series on the RHS of 10 or so should do a few decimals
@Semiclassical point P is on the terminal side of angle theta, P(0,5) is the given point
Is there a way I can do this?
I cant make that into an right angle
because I don't know how long the hypotenuse will be
15:49
Hi @MikeMiller
Welcome back to the realm of the living.
Is this it? I always confuse the two realms, they look so similar.
@BalarkaSen is there a difference?
MM came to the town today. Talked to him about a bunch of stuff
16:05
Hello! :) Anyone got any ideas to find all continuous function that satisfy the Jensen inequality, ot specifically
f( (x+y)/2) = ( f(x) + f(y) ) / 2
@MikeMiller He said something along the lines of it being open whether any surface bundle over surface is hyperbolic (constant negative sectional curvature). Is this it?
@Balarka The big open question is whether that's flat, but flat doesn't mean what you think it means.
@BalarkaSen MM?
Marshall Matters?
@Krijn the one above you
@DHMO Then why would he respond to Mike with "He said"?
16:17
he was talking to me
@BalarkaSen A connection on a fiber bundle is a distribution of codim (dim F) on the total space that transversely intersects the tangent spaces of the fibers. It's flat if this is an integrable distribution. Every bundle over a circle is flat (exercise), but the Milnor-Wood inequality proves that there's only finitely many flat circle bundles over a surface. It's open whether or not all surface bundles over a surface are flat.
There are examples of non-flat surface bundles when the dimension of the base is 6 iirc.
@MikeMiller Ah
Actually, I think you're right, and it's also open whether surface bundles over a surface can be hyperbolic.
are you two up for a definite integral challenge?
I see, got it. Reading your explanation of flat.
16:22
No.
ok
Me neither. Also, I wasnt talking to you. MM is someone else
I see
Oh, right, I meant "whether there is any".
That flatness condition is the same as having a foliation the leaves of which intersect the fibers transversely, yeah?
Yes, which then implies that the projection to the base is a local diffeomorphism.
Yeah, the distinction in the two cases is "are all of them flat?" and "are any of them hyperbolic?" Both are open, and the mapping class people are working on both. I dunno how much success there's been.
16:30
32
Q: A community project: prove (or disprove) that $\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ is convergent

Jack D'Aurizio As the title says, I would like to launch a community project for proving that the series $$\sum_{n\geq 1}\frac{\sin(2^n)}{n}$$ is convergent. An extensive list of considerations follows. The first fact is that the inequality $$ \sum_{n=1}^{N}\sin(2^n)\ll N^{1-\varepsilon}\qquad\text{or}...

So... does anyone know I good convergence accelerator to apply to the above sum?
@MikeMiller Err. The trivial surface bundle on the circle is flat, but the projection to the base is clearly not a local diffeomorphism...? Oh you mean for a surface bundle over a surface
I should have said this:
are you two up for a definite integral challenge? It's definitely integral to definite integrals.
@BalarkaSen The projection of the leaves.
Ok, ok, got it
obe
obe
16:44
hi guys what is a good vector analysis book that is like abbott's understanding analysis book that covers multivariable analysis in the same way?
hi chat.
@Semiclassical are you up for a definite integral challenge? It's definitely integral to definite integrals.
You might as well give the integral. Not sure I'll be interested, though.
$\displaystyle \int_1^2 \int_1^2 \int_1^2 \int_1^2 \dfrac {x_1+x_2+x_3-x_4} {x_1+x_2+x_3+x_4} \ \mathrm dx_1 \ \mathrm dx_2 \ \mathrm dx_3 \ \mathrm dx_4$
Eh.
Well, the value can't depend on which of the four terms on top gets the minus sign.
16:54
does anybody know what $X^2$ might mean if $X$ is a subset of a topological space?
and it doesn't mean $X\times X$
@JorgeFernándezHidalgo mas contexto?
is there another possible interpretation?
Second, the integrand is of the form $1-2x_4(x_1+x_2+x_3+x_4)^{-1}$.
apparently it was in someones topology 1 exam
he was asked to consider the subsets of a space such that $X^2$ is finite.
So therefore I can symmetrize the integrand to $1-\frac14 (2x_1+2x_2+2x_3+2x_4)(x_1+x_2+x_3+x_4)^{-1}=1/2.$
16:56
@Semiclassical wonderful.
you're definitely talented
It generalizes to arbitrarily many such variables, of course.
If you want other integral challenges, there used to be a poster here who would do a lot of them.
we just did many in the h bar
chris's sister?
Ah.
yeah.
Hi there
I got a question
The answer to this question, holds for Hilbert spaes?
16:59
@Semiclassical you can see it starting from here
spoiler alert, and warn you that there are many questions

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