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00:15
Idealism says there aren't problems; cynicism says there aren't solutions
Hello @Semiclassical
I've once heard someone say that the media is "the rich telling the middle class to blame the poor." Thoughts?
That someone sounds like they have they're own agenda they want you to believe in.
@AkivaWeinberger I once heard someone say that social media is "the middle class telling the poor to blame the rich." or something along those lines
How does that make sense
00:21
I don't think social media is usually politically/economically motivated.
and when it is, it is almost always done by big players.
:-) well, I heard it from someone who doesn't use social media
Anyway if I wanted to understand something as complicated ad the role of the media in society, I'd probably look for something beyond a neat sounding catch phrase,
In perhaps my favorite novel ever (Sometimes a Great Notion) , there is a character who comes up with these kinds of quips in some hope for understanding mankind. It doesn't go to well for him.
Oh yeah! If anyone likes time travel, watch Stein's gate.
@Simply That show really was fantastic.
00:35
is there a site like imdb
for books?
everything I find is behind a paywall.
Grimaldi, Discrete and combinatorial mathematics, 5th edition, Addison-Wesley
Can someone tell me if there is a pdf on this book ?
I could not find one
@Kasmir did you look on libgen?
@PVAL-inactive I dont know what that is =p
libgen.com ?
I don't know what the precise domain is and I won't try and figure out on the comp. I am on.
That is generally the best place to obtain textbooks in a morally ambiguous fashion.
I don't understand really but thanks :)
I would buy it if they ship it fast
But it takes month or so and I need it soon
00:45
If this is for a class, a standard procedure is to place a copy on reserve at your university library.
So you can see if anyone is using that.
Okay will do :)
the books in swedish are easy to find
but most of the books in english we have to buy from amazon or if lucky find old copy
01:08
Oh libgen is a large source of my meme (=textbook) stache
@Daminark Hi Amin
How's it going @Kasmir?
Just had exam on complex analysis =p it went good :D
what about you ?
all i got left this year is to pass combinatorics and i need that book ><
Good luck!
Hey, I'm trying to look up some information, but don't know what to search for. I'm looking for something like the PDF of sample mean, only for sample maximum.
If I take the maximum of 5 samples, what would the PDF look like. Is this a standard math concept?
01:34
... I guess its called an "order statistic"
 
1 hour later…
user278130
02:43
Hi. Can someone please take a look at this question? math.stackexchange.com/questions/2242814
03:47
hi @arctictern
I wanted to ask the following question I am solving from Michael Atiyah. Let A be a ring, I be an ideal, M an A-module. Show that $(A/I) \otimes M$ is isomorphic to $M/IM$. The way I solved it was consider the following sequence $0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0$ this will induce the following exact sequence $I \otimes M \rightarrow M \rightarrow A/I \otimes M \rightarrow 0$. To finish the proof I want to show that $I \otimes M \cong IM$. I showed that the map
$a_1 \otimes m \mapsto a_1m$ is well defined and surjective.
I am having troubles with injectivity though
Hey everyone!
hi @Daminark
How's it going?
Hey @Mike!
04:00
good just solving problems preparing for final
Suppose $A=\Bbb Z/4$ and $I=M=2\Bbb Z/4$. Then $I\otimes_A M\cong M$ but $IM=0$.
Lol, I'm doing a pset
I know it's probably been a little while since you've done calculus, but is it true that a homeomorphism with one direction smooth is a diffeomorphism? I'm inclined to say not but it'd be nice if this were true...
@arctictern maybe should I prove directly that kernel $(M \rightarrow A/I \otimes_A M)$ is $IM$ ?
iunno
I am sick in bed at 11pm
oh okay sorry to hear that :S
04:04
Hope you feel better @arctic
Dammit that isn't true... :(
@arctictern I got it. So the map $I \otimes M \rightarrow M$ works as follows $i \otimes m \mapsto im$ so first map has image $IM$ so we are done.
@Daminark that is not true
Yeah
So map something smoothly that has a cusp into something that doesn't have a cusp then one direction is smooth
but other direction isn't smooth
I think square --> circle works
04:26
This is kinda merp
05:02
@Daminark (using your legal name for ping): No, there's a very famous homeomorphism that is smooth but whose inverse is not even everywhere differentiable. Think most basic precalculus.
@arctictern: I hope you feel better soon. I have been sick for almost a month now. :(
@Ted Early in my grad career I wrote a proof that every twisted sphere (including all the exotic spheres in higher dimensions) was smoothly homeomorphic to the standard sphere using a version of stuff Bing did.
I've since realized that $x^3$ kills it
That was to what I was referring :P
Oh, interesting, @PVAL. I had no idea there was any differentiability there.
I abandoned that line of argument and made some loose argument in its place
Because I have a paper to write, still
@Ted There's a MO answer by Greg Kuperberg that probably is a more direct proof, but it always seemed like a pain (for me) to make rigorous.
05:07
@PVAL: I admire Hatcher for his exercises. I won't say the text itself is the best, although I am disappointed in most alg. top. texts, tbh.
Interesting, PVAL. But Greg is brilliant about most everything.
I sort of decided (with input from my adviser) though that thinking about that kind of stuff was a bad direction and go to in
The problem says to show that if the boundary of a compact, convex set is a manifold, then it's homeomoprhic to the sphere. The idea is to push the set so that the origin is inside, and then surround the set in a large sphere (radius $r$) and push out by $f(x) = r\frac{x}{\|x\|}$.
Sort of like Stallings's paper "How Not to Prove the Poincaré Conjecture"?
Now, convexity gives us that this is a bijection, and it's clearly continuous
Since the boundary is away from the origin, we get smoothness as well, and compactness gave us a homeomorphism
Homeomorphism is easy. Diffeomorphism is far from easy.
05:09
The wishlist proposition was that this was enough to say diffeomorphism
@Ted I just decided it was a bad idea because people weren't actively thinking about similar stuff.
I'm not sure I know how to do diffeo, although Spivak assigned that to us when I took graduate manifolds/geometry from him in 1974.
So if I committed my grad school to it, then I'd feel like it'd be hard to continue in math.
Loosely speaking I just said that this map in general has rank one less than the dimension of the space, which is in fact the dimension of the sphere and of our manifold, so that should suffice to say inverse function theorem locally. Since it's a bijection, that's enough to say total diffeomorphism
Oh, you said convex. It's true even for star-shaped. But then it gets harder.
05:10
Oh I didn't realize you took a class under Spivak
No, what you're sketching doesn't look very valid to me.
@Ted Homeomorphic is very easy in either case.
I agree, @PVAL.
Is there a way to use convexity?
Where does this break?
05:11
Probably a very wise decision to change path, @PVAL.
How do you get the hypotheses of the inverse function theorem at every point, Demonark?
@TedShifrin Yeah. Now that what I see people are posting as preprints and have some experience to evaluate them myself.
You don't even know the boundary is piecewise smooth, Demonark.
I feel like I wish I would have tex'd it up and posted.
It is assumed to be a smooth manifold
You just said boundary of a compact convex set.
05:12
and probably will do that and post it.
I said if the boundary of a compact, convex set is a manifold
I saw that question on main
Oh, I missed that, Demonark. So in that case you can probably prove distance to the boundary is smooth, applying convexity somehow.
Yeah
But I don't know that immediately.
05:14
If it's a smooth submanifold, I think you just use distance squared
or distance
which is automatically smooth away from the origin
Oh, sure I do.
Right, @PVAL.
since its smooth on R^n
away from the origin.
Oh, and we use convexity to prove each ray from the origin intersects precisely once.
That will fail for star-shaped.
nah it still works
The difficulty is showing diffeos for star-shaped sets where the boundary isn't a smooth manifold
i.e. showing the interior is diff. to \Bbb R^n
Assuming manifold you mean? The problem Spivak assigned us was for any star-shaped compact set, I'm pretty sure. I can look in his book.
OK.
05:16
Hey @Alessandro!
Oh, it must be Alessandro's morning.
Hi @Dami @Ted
@Ted If the boundary isn't smooth then it doesn't really make any sense.
It is my morning as well :P
Lol jk
05:16
@PVAL: I think the problem was to prove the open set is diffeo to a ball. I'll look it up.
It's not morning until the sun rises if allnighter, or otherwise when the alarm goes off to wake up :P
I'll take a bet that Spivak didn't know how hard that was.
6
Q: Is there a citeable reference for star-shaped open subsets of R^n being diffeomorphic to R^n?

Dmitri PavlovA folk theorem says that star-shaped open subsets of R^n are diffeomorphic to R^n. Is there a citeable reference for this result? For the sake of being definite, let's say that “citeable” means indexed by Mathematical Reviews or Zentralblatt, or available on arXiv. (The answer https://mathoverf...

30
A: What are the open subsets of $\mathbb{R}^n$ that are diffeomorphic to $\mathbb{R}^n$

Georges ElencwajgAd question 1): Yes, all open star-shaped subsets of $\mathbb{R}^n$ are diffeomorphic to $\mathbb{R}^n$. This is surprisingly little-known and there is a proof due to Stefan Born. You can find this (fairly complicated) proof in Dirk Ferus's course notes http://www.math.tu-berlin.de/~ferus...

Oh, no, he just gave homeomorphic, @PVAL, but then gave a picture (which I was remembering) where the obvious construction fails miserably.
The second question is the one I meant to link
The answer there uses Whitney extension.
which is HARD...
and its still a really non-obvious construction from that.
I remember learning Whitney extension in my dynamical systems class first year in grad school. I thought it was so f***ing awesome.
Anyhow, Spivak just asks homeo. I'm still not sure I remember how.
Convex is easy.
05:20
you still do the obvious thing
and project away from the star point.
The obvious thing fails.
@Ted Do you recommend doing dynamical systems at some point? It sounds like a neat subject
okay, then whats the counterexample?
It's amazingly neat, Demonark. For a warm up for undergrads, see Hirsch-Smale's beautiful book.
Spivak draws the picture of a disk with a hair growing in from the boundary along the radius, @PVAL (part way only to the origin, of course).
Will do. I think in the RTG bootcamp we're gonna do some but probably not much
(Maybe I should talk to Amie Wilkinson)
Ah, right, I remember Hirsch-Smale!
A few of us were using that with Perko for ODEs, I also like it. :)
05:22
@Ted That isn't starshaped to me.
Sure it is, @PVAL, relative to the origin.
Then I don't understand your construction.
Distance from the origin to the boundary is very discontinuous.
@Alessandro: Did you do OK on your probability exam?
It'll begin in an hour more or less @Ted
Star-shaped (according to wikipedia) means that there exists an x_0 in your set where the line segment between x and x_0 is always in your set.
05:24
Oh, well, do well. I've got confidence in you. :)
Yes, @PVAL, my $x_0$ is the origin.
("with respect to")
ahh I see
"Question 1: Find your expected score"
You have like a complete singularity along the line.
well, the distance to the boundary function, as a function of $\theta$ is $1$ everywhere except one value, where it's $1/2$, say.
@TedShifrin Thanks, it's definitely not my favourite subject, but I feel fairly read
05:26
I actually loved learning/teaching it, @Alessandro, but you and I have very different tastes. :)
Lol I'm still conflicted on probability
I would expect nothing less from you, Demonark.
That's fair, conflicted on just about everything
Well, then your variance is small.
Except maybe finite group theory
05:28
By the way, an analysis question motivated by probability that I was discussing with Balarka yesterday. Supposw that $f:\Bbb R\to\Bbb R$ is analytic in a nbhd of $0$,what kind of conditions do I need to ensure that the power series will converge in a disk if I plug in a complex variable instead?
That's so far been uniformly fun
@Alessandro: Unless I'm being stooopid, no condition.
@Ted It's still easy you get a surjective map to the ball just by going proportionally along those rays.
@PVAL: Does "map" mean continuous?
Then your non-trivial inverse sets are finitely many lines.
05:31
Ok, thanks, Balarka agrees with you, while I have no idea, we just started complex analysis
and that's easy to make into a homeomorphism.
Sorry, @PVAL, I'm not seeing why it's continuous.
@Alessandro: It's a reasonably natural and standard question.
Heya @MikeM
maybe it isn't
whatever homeomorphisms died like 30 years ago
LOL, huh?
with bing
05:34
Oh :P
Hey @Mike!
What's bing?
I was wondering about it while thinking about the moment generating function and the characteristic function of a random variable, because if I have a series expansion for the first it seems like I can just plug in $it$ for $t$ and get the second @Ted
Yes, @Alessandro, that's a common game, I think.
What you know about convergence of (real) power series will go over immediately to complex power series — think about the proof of the ratio test, root test, whatever.
Demonark: A yummy cherry.
I did not think cherries would kill homeomorphisms
shrug
R. H. Bing was a very famous old-style geometric topologist who trained many of the topology minds of the past half century.
05:39
Fun fact: R.H. was exactly his name, not just the initials
Oh that's neat!
I wasn't really referencing Bing's death, but that works too (given my location).
Just not too much people thinking about that stuff since I guess he died and lots of his students retired/moved on to different research/focused on teaching.
Anyhow, @PVAL, I think I (perhaps with some hints) figured this out in grad school, but that was a few days ago.
I have to go, I'll be back after the exam, bye everyone!
Good luck, @Alessandro.
05:42
See you @Alessandro! I should do a TAPS paper and either sleep or work on analysis now, so see the rest of you around as well!
I think the only thing that can happen is you have an entire line in the sort of discontinuity of the thing
I guess the point is there is some ball around the star point
Agreed.
and you can project the boundary onto that ball
Not one-to-one, but granted.
Yeah but it will have finitely many inverse sets
which are all line segments
05:45
Why only finitely many hairs allowed?
or maybe not.
I don't see why you can't have a countable collection.
I'm not sure why this is revelant, though.
There's lots of things you can do to show maps with certain inverse sets can be approximated by homeomorphisms.
Nearly all of this is due to R.H. Bing and his adviser.
Oh, cool. I know I wasn't that smart in grad school, though.
yeah but I am
05:48
:)
I would imagine though that you might have known that if you identify all points on a line (and nothing else) in $\Bbb R^n$ you still get back $\Bbb R^n$
Line SEGMENT
Yes, that is plausible.
If you take a ball and then project every point onto the ball that maps continuous on R^n
so of course its continuous on your star-shaped thing.
So take a ball around your star point
Yes, i agree the projection is continuous.
do it to that ball and you get a surjective map where your killing some line segments
05:55
Well, wait, you're projecting to the boundary (and then extending with the identity map)?
No, the projection I'm thinking is killing lots.
fine
idk
Well on the boundary I'm pretty sure its only killing these sort of line singularities
you get
so the boundary is probably homeomorphic to the sphere
I don't think this is quite what we want. I'll think about it some more tomorrow.
06:23
Well considering you guys were talking about Bing I'm going to ask a low dimensional topology question
Is every subcontinuum of a one dimensional locally path connected continuum locally path connected?
(covering dimension, that is)
I asked this on MSE but it only got like 3 views and it's been flooded out haha
 
3 hours later…
09:46
If $F$ is a field and $a \neq 0$ is a zero of $f(x) = a_{0} + a_{1}x + · · · + a_{n}x
^n$ in $F[x]$, Then $1/a$ is a zero of $a_{n} + a_{n−1}x + · · · + a_{0}x^{n}$.
I thought of this a s applying continuously inverse of $a$ $n $ times
10:33
To prove it, notice that the latter polynomial is $x^nf(x^{-1})$
yup,actually
2
Q: Is $1/a$ a zero of $a_{n} + a_{n−1}x + · · · + a_{0}x^{n}$ ?

BAYMAXIf $F$ is a field and $a \neq 0$ is a zero of $f(x) = a_{0} + a_{1}x + · · · + a_{n}x ^n$ in $F[x]$, Then $1/a$ is a zero of $a_{n} + a_{n−1}x + · · · + a_{0}x^{n}$. I thought of this - As $a$ is a zero of $f(x)$ so $f(a) = a_{0} + a_{1}a + · · · + a_{n}a ^n = 0$ Now as $a \in F$ so inverse of ...

if splitting field $K =Q(a ,a\omega,a\omega^2)$
then it is equal to $Q(a,\omega)$
How we get second line from first line
I think it is something related to basis but thow?
10:52
Is there a known relationship between the change in length of a chord and the change in measure of the arc it intercepts?
11:28
0
Q: confused in integration

search For the integration I tried different substitution but they all are useless . And I am not getting any other method . Can anybody provide me a hint

11:43
@BAYMAX A chord will be given by some line, so set the equation of a circle equal to the equation of a line, and use the endpoints to compute the arc length
off the top of my head
0
Q: Gauss' theorem over sphere

LozanskyDetermine $$\iint_S \textbf{A} \cdot d\textbf{S}$$ where $$\textbf{A}(x,y,z) = (xz^2,x^2y,xyz)$$ and S is the sphere $(x-1)^2+(y-2)^2+(z+3)^2=4$ Attempted solution Let's use Gauss' theorem. We obtain $\nabla \cdot \textbf{A} = x^2+z^2+xy$. A parametrization of the sphere is given by: $$\Phi(r,\...

12:02
yes@Excalibur42
12:35
$x^p + a$ in $\mathbb{Z}_{p}[x]$ is not irreducible for any $a \in \mathbb{Z}_{p}$
many questions like this in main I find them hard?
13:05
That's because $x\mapsto x^p$ is an automorphism of Z/pZ (known as the Frobenius endomorphism) so something will be mapped to the inverse of a
any simpler version?
any resources where this is dealt with,I could not find it any where in the web except some MSE posts ?
automorphism then is it this one $f : Z/pZ \rightarrow Z/pZ$?
@AlessandroCodenotti
Automorphism means that it's a field isomorphism from Z/pZ to itself
ok
Lets start from scratch?
We have to show $x^p +a$ is irreducible
in $Z_{p}[x]$
Oh, much easier, $x^p=x$ in Z/pZ
for any prime $p$
elements of $Z/pZ$ are of the form $a + pZ$
$a \in Z$
is this correct @Alessandro
13:12
Yes, but it's easier to just pick the representatives 0,1,...,p-1
actually i was thinking why$ x = x^p$ in $Z/pZ$
Fermat little theorem
oh yes $x^{p-1} $ is congruent to $1 (mod p)$
so the problem reduces to $x + a$
so we need to show is $x + a$ is irreducuble in $Z_{p}[x]$
also $a \in Z_{p}$
so there are chances that $x + a$ will be zero
so then it implies it has a zero
which implies reducible
@AlessandroCodenotti
That's not the same, $x+a$ is irreducible (it's linear), $x^p+a$ is not
But knowing $x^p=x$ is helpful to find roots of $x^p+a$
so how @AlessandroCodenotti
so we cannot replace $x^p$ by $x$
?
13:20
It's not the same polynomial
ohh
ya
Hello dear fellow, I asked a question yesterday but had no luck as of yet : maybe someone here would be kind enough to give me a hint ? math.stackexchange.com/questions/2242005/…
so now we should use $x = x^p$ to find the roots of $x^p + a$
Right
One in particular should be easy to find
1?
yes
13:23
Why should 1+a be 0?
as a can be p-1?
$a \in Z_{p}$
You have to find one for every $a$
ohok
yes it was for any a
also the root cannot depend on $a$?
Why can't it depend on a? As long as you find a root for every a you're fine
$ap -a $?
13:28
ap is 0 in Z/pZ
and $(-a)^p = -a$ thus -a +a = 0
so $ap -a$ is a root of $x^p + a$
Yes but ap-a=-a
yes
13:30
No need to carry ap around
oh
so $-a$ is the root of $x^p + a$
and $-a \in Z_{p}$
so it is not irreducible in $Z_{p}[x]$
is this correct@AlessandroCodenotti
now another thing is does not irreducible means reducible ?
I'd say $-a$ is a root rather than the root, since there could be as many as p
(That won't happen though)
@BAYMAX yes
13:33
If $|w|=2$ then what is locus of z=w -(1/w)
can anybody help me in this
Thanks @AlessandroCodenotti
@user123733 I think its a disc
not sure though
I go like this
@BAYMAX in my book it is given as an ellipse with eccentricity 4/5
but I could not understand how
Consider $|z| = |w - 1/w| \leq |w| + 1/|w| = 2 + 0.5 = 2.5$
$|z| \leq 2.5$
I dont know what is wrong in my argument,but I use this and create troubles to me.
Yeah I hope you get a nice answer
@user123733
hey@AlessandroCodenotti
any help on this
3 hours ago, by BAYMAX
if splitting field $K =Q(a ,a\omega,a\omega^2)$
@AlessandroCodenotti can you help me in that
No, sorry @user123733
13:42
its fine
I'm not sure what you're asking @BAYMAX
@user123733 you can initiate something?
Like why it was sufficient to take $Q(a,a\omega)$?
@AlessandroCodenotti
How many Hamiltonian circuits are there in a complete graph with 6 vertices?
What are $a$ and $\omega$?
actually what is the splitting field of $x^3 -2 $ over $Q$
@AlessandroCodenotti
$a$ is a real root of the polynomial
$\omega$ is the cube root of unity
13:59
@TedShifrin "Congratulations, you are a grammar master! You have a superb understanding of even the trickiest grammar rules. Not only do you know the difference between affect and effect, but you also never confuse your tenses. You must be an English scholar because only 4% of Americans can get a perfect score on this test." XD
0
Q: complex number with determinant

user123733Let $z_1$ and $z_2$ be two distinct complex numbers and let $\,z=(1-t)z_1 +tz_2\,$ for some real number $t$ with $0<t<1$. Then we have to prove $$\begin{vmatrix} z-z_1 & \overline{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1} \end{vmatrix}\;=\;0$$ I thought about it, but don't ge...


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