« first day (596 days earlier)      last day (4426 days later) » 

leo
3:01 PM
hi there
 
hi leo
 
leo
given any nonmeasurable set $V$ it is true that if $E\subset V$ is measurable then $m(E)=0$?
 
Of course not.
 
leo
I'm able to prove this for Vitali sets
 
For those it holds.
Else you can take a measurable set $E$ of positive measure disjoint from $V$ and consider $E \cup V$.
 
leo
3:05 PM
yes
 
This set isn't measurable but it contains a measurable subset of positive measure.
 
leo
you are right
My situation is this: $I$ an interval, $V$ a Vitali like subset of $I$, I want to prove that if $E\subset I\setminus V$ measurable, then $m(E)=0$
 
How is $V$ constructed?
 
leo
by considering the equivalence relation
$x\sim y$ iff $x-y$ is retional
I mean rational
 
(you can edit your posts by hitting the up arrow on your keyboard)
So it is a Vitali set?
 
leo
3:10 PM
yep
 
(\sim = $\sim$)
 
and did you prove that it has full outer measure?
 
leo
I have proved that the outer measure of $V$ is positive
 
Nice.
 
But you need to do some additional work in order to guarantee that it has full outer measure.
 
leo
3:12 PM
$I\setminus V$ has positive outer measure as whell
I don't know what you mean by "it has full outer measure."
 
Outer measure $1$.
 
leo
ah ok
So, my $V$ has outer measure $1$
 
Does it?
without doing some work you can choose your representative set for $\sim$ in $[0,1/2]$ and of course then the outer measure will be at most $1/2$
Since you didn't tell me what exactly $V$ is, I can't help you, yet :)
 
leo
oh sorry
@tb pick an element of each one of the equivalence classes determined by $\sim$, put all them in a bag, $V$ is that bag :-)
 
But how do you guarantee that $V$ has outer measure $1$ this way? As I said, nothing keeps you from picking only those in the $[0,1/2]$ half of the bag.
 
3:19 PM
@tb I don't see how I can conclude $z^n$ from this. Are you sure this works?
 
leo
@tb at first glance, nothing
 
You want to construct a Vitali set with outer measure 1?
 
@leo see this thread
 
leo
@JonasTeuwen yep
 
@MattN What are the continuous homomorphisms $\mathbb{R} \to \mathbb{C}^\times$? Yes I am sure it works
 
3:21 PM
Right, I've asked such a thing before as @tb has just linked to.
Robert Israel posted a solution on sci.math a long long time ago.
A long long time ago. I can still remember like the ... whatever I forgot the text.
 
leo
@JonasTeuwen thanks
 
That was correct (if you restrict the target to $S^1$). Now you need to guarantee that this homomorphism descends to $S^1$ (since it came from there after all) by factoring over its kernel, so the kernel must contain $2 \pi \mathbb{Z}$. Now think about what closed subgroups of $\mathbb{R}$ contain $2 \pi \mathbb{Z}$.
 
No it was wrong, I think. They are $\varphi : \mathbb{R} \to S^1$ , $x \mapsto e^{ixy}$ for $y \in \mathbb R$.
 
Knu
Hi, how would you type 5n twice? (5 and 10 but not 15)
would 5n x 2 be correct?
some kind of repeat argument if you want
 
Okay, I think this should be enough for the moment, I should do some serious work.
 
3:29 PM
@tb Yes. Thank you and see you later!
 
leo
what means JDH by " It follows easily that V is not measurable and has inner measure 0, since otherwise the translates modulo 1 would have infinite measure inside [0,1]"
follows easily that $V$ is'nt measurable
that's true
got it
:-)
 
Knu
I don't want to use a limit which is less precise. And googling nth gives me nothing relevant.
 
leo
no problem
 
Knu
How would you express that repeat argument? Would a multiplication be correct or it would mean something else?
I guess no one cares.
 
3:45 PM
@Knu what do you mean type 5n twice?
 
@JonasTeuwen Are you there?
What does "a homomorphism descends to $S^1$" mean? Is this a fancy way of saying it maps to $S^1$?
 
4:13 PM
@MattN Yes!
And now I'm a goner! Good night ding ding ding ding.
I'll go.
Back soon.
 
leo
if $m_i$ is the inner measure defined as here it is true that $m_i$ is at least finitely subaditive
 
Knu
4:32 PM
@paul I was trying to express 5n twice (with a repeat argument) without using a range nor the result (5 and 10).
But after some research it seems the notation is not intuitive enough.
Unless you come up with something I didn't read.
 
it's a fact that the lie algebra of an abelian lie group is also abelian. can you show this using pushforwards? specifically using the pushforward of the inversion map (which is a group homomorphism)?
 
Knu
@paul {5*n | n in [2]} that's what I found. Do you have anything better?
 
@MattN I'm back. I'm melting.
 
I sincerely hope @MattN is doing extraordinarily well. I wonder why Matt drops in and out, so many times... Are you kinda busy now? @MattN
 
Knu
The stared imgur is for real? It's not photoshoped?
 
4:42 PM
definitely photoshopped
it's making fun of sensitivity to islam
 
but what do you mean "express 5n twice". I don't understand what you mean? Have you got the explicit original question?
 
does anyone know some manifolds/lie algebras? i am just looking for a bit of direction in what i think should be an easy problem
 
@Knu $\{5n \mid n=1,2 \}$ is just the same as writing $ \{5, 10 \}$
 
I am fairly certain that someone knows some manifold and Lie algebra material. I am not sure if someone in the channel knows, though.
 
@Eric: What makes you think it's photoshopped? That means the actual words on the Fox news screencapture were altered, not that words were added onto the side.
 
4:59 PM
it looks photoshopped, for one. two, it is an obvious parody of both fox news and the response to the incident
 
hm. okay.
 
wat :O
 
it is easy to troll partisans, apparently
 
@EricGregor and what does it prove?
 
5:07 PM
@Ilya, what does what prove?
 
@KannappanSampath I'm doing fine thanks. And yes, I'm busy! : )
 
@JonasTeuwen : )
 
@EricGregor you told that it is a photoshop and then put a screenshot. For which reason?
 
@Ilya, to show how easy it is to photoshop
 
5:12 PM
@EricGregor I think no one here has doubts
 
anon seemed to think the original was real
 
I agree with anon, that's why I've asked you what should your picture prove. If it is easy to photoshop it doesn't say anything about the original
 
did you see the last link i posted?
you guys are gullible
with all due respect
 
@Ilya It's clearly a photoshop to troll, as Eric has amply shown.
 
@Eric: I saw it, and?
 
@anon I need some TeXnical help on the main site. Can this be beautified? Please help me.
 
@EricGregor do you think I will learn Dutch in 5 minutes?
 
there is no Fox News "watermark" on the right of the original, for one. second, it assumes a kind of monstrous stupidity of Fox News that only exists in a liberal caricature
google.translate.com
 
thanks for the translation. So you rely upon DeMorgen
 
@Kanna: I don't see what the issue is with your answer.
 
5:18 PM
@anon Does it not look ugly? The fractions...
 
@Ilya, and common sense
 
Looks fine to me, Kanna.
 
@Ilya, i just looked at your profile, it says you live in the Netherlands! how do you not know Dutch?
 
Because many people in The Netherlands don't speak Dutch.
 
5:25 PM
@EricGregor I appreciate your attention to myself, but I didn't learn it
 
@Jonas, really?? i didn't know that
i though Dutch was the national language
 
Yeah, people actually speak Spanish.
(is joke)
 
@Jonas: is DeMorgen reliable? ;)
 
@Ilya The newspaper?
 
@JonasTeuwen yes. The Spanish joke is thin
 
5:26 PM
@Ilya It is something like NRC in NL.
@EricGregor No but if people notice you don't speak Dutch very well they will speak English.
 
wow, i didn't know that
that seems suicidal for a country. i imagine holland won't stay dutch very long
btw, to give this conversation a chance of becoming mathematical, do either of you guys know basic manifold/lie algebra theory? i have what i think should be a rather straightfoward question about lie algebras of lie groups that has been bugging me.
 
You might as well just ask your question. The only thing I know about lie algebras/groups are their definitions, though.
 
Broke 5k! : )
 
the lie algebra of an abelian lie group is also abelian. this can be seen using basic representation theory. can you show this using pushforwards? specifically using the pushforward of the inversion map (which is a group homomorphism)?
 
What do you get at 5k?
 
5:32 PM
@anon Tag wiki edit can be approved! That's all (officially.) Unofficially, I am satisfied! : )
 
@EricGregor what do you mean?
 
I can approve edits!
 
Reviewing edits is a thankless task. Other than the badge you get for 100 actions on them.
 
Right. I'm happy with my recent answer you know!
 
@Ilya, Lee suggests that you can use a) the fact that inversion is a group homomorphism, and b) that the pushforward of inversion is $i_*X=-X$, to prove the claim
i don't see how to connect this to connect these ideas to the lie algebra itself. we want to show that the lie bracket is trivial, i suppose
 
5:37 PM
@Eric what if you do the inversion on a product and then take the differential? (I'm out of my depth here.) | Also if you check Ilya's reply link you'll see he was replying to your "suicidal" comment.
 
@anon what do you mean precisely by "taking the differential"? the pushforward of the inversion on the product?
 
yes, that's what I mean.
 
@anon how would that connect to the Lie algebra?
@Ilya, regarding my comment, i suppose if you don't find it obvious that cultures die when languages die and people become insufficiently insular, it would take quite a bit of discussion. i am happy to have the discussion but i don't want to make this a political forum.
 
I'm sort of shooting in the dark. | Also, you referred to "country" rather than "culture" previously.
 
i said "i imagine holland won't stay dutch very long". they may remain The Netherlands, but the Dutch culture and people won't survive, that's what i'm claiming. of course you could just rename the new people and language Dutch, but that's cheating
 
5:43 PM
"This is a strange dialogue" writes, The Chaz, a M.SEr....(for those who did not know.)
 
@anon, i'm sure your idea is the right one, i just can't find that explicit connection. i just need to go back to the definitions, i'm kind of new to this stuff
 
@EricGregor where are you from if I may ask
 
Hi @MarianoSuárezAlvarez
 
@Ilya the American Northwest
 
5:45 PM
@EricGregor but you know Dutch?
 
if $\varphi$ is the inversion map then $d (\varphi\circ\exp) X=-X$, is that what your note (b) is saying?
 
no. but i know how to use Google translate :P
 
icic
 
@anon as long as your $d$ is the pushforward: $d: T_e G\to T_e G$
 
5:48 PM
oh man, overload on the \mathrm
 
@anon nope, \text
he is just wasting ink
 
A closing related issue:
I contend that the latter question can be merged with the previous one on all grounds.
What do we have to say?
 
You linked to the same question twice.
 
i have to go to lunch. if anyone sees my question and feels like helping me out, please feel free to send an @ to me. hopefully i'll figure it out over lunch.
bye
 
@KannappanSampath closely?
 
5:53 PM
@anon @anon Edited.
@Ilya Pun?
 
@KannappanSampath if only I knew what does Pun mean
 
@Mariano Your opinion is useful too.
 
It should probably be closed/merged.
 
But, how do we merge?
 
only mods can do that.
 
5:59 PM
So, flag it for a mod, then?
 
nah. just vote to close or something.
I don't think I've even seen a merge happen before.
 
OK. I'll vote to close the second one as a dupe of the first. I'll leave a comment there.
 
so... the unit sphere tangent bundle to S^2 is a projective space. anyone can tell me what the homeomorphism is?
 
@MarianoSuárezAlvarez identify RP^3 with SO(3) via quaternions and the cover SU(2) -> SO(3). the first column of a matrix in SO(3) gives you the base point, the second one gives you the direction in the unit tangent bundle.
 
6:15 PM
Ahh
thanks
stupidly, I started with the Stiefel manifold :)
 
Euler angles would probably offer another way to put it, I think.
 
SO(3) \cong SS^2 was the one map missing
i like to see SO(3) \cong RP^3 as follows: a positive rotation gives you an axis and an angle
that pair is a point in the closed ball of radius \pi in R^3
(in "polar coordinates")
but one has to identify rotations of angle pi and -pi
so one gets RP^3 by identifying antipodal points on the boundary
 
That's nice!
 
one avoids quaternions that way :)
(which is of course bad!)
 
heh :)
 
6:40 PM
@tb Something is wrong with the line of gravatars for people in this room.
@MarianoSuárezAlvarez I closed my browser and shut my computer down for 4 hours and when I came back my gravatar was still in the line.
 
Heya !
 
Does anyone have the link to different ways to evaluate the gaussian integral?
I rember seeing it on the site, but can not recall where
 
Hello!
 
6:55 PM
30
Q: $\int_{-\infty}^{+\infty} e^{-x^2} dx$ with complex analysis

JasonMondInspired by this recently closed question, I'm curious whether there's a way to do the Gaussian integral using techniques in complex analysis such as contour integrals. I am aware of the calculation using polar coordinates and have seen other derivations. But I don't think I've ever seen it done...

@N3bu this one?
 
Free at last. I was stuck in the gravatar line for over 4 hours.
 
14
Q: Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

JichaoHow to prove $$\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$

or this one
 
@anon There was another one where one recalled the integral was sillimar to a cylinder or something. A method I had never seen before.
It is perhaps the second one, I will have a look.
=)
 
second answer on the second one
 
thanks
=)
 
7:08 PM
Free, free, free at last... I was stuck in the gravatar line for over 4 hours.
 
7:29 PM
@Skullpatrol can't see that. It is blocked.
 
@Skullpatrol same thing
 
@anon has that question not been answered before?
 
dicks
 
7:34 PM
@anon yeah
 
well, it gave me a thumbnail
 
Who is EMI?
 
It might be a campaign to get EMI to do something.
 
@robjohn Thanks.
 
Hi @robjohn
 
7:37 PM
@KannappanSampath hey there. What's up?
 
@robjohn Can you youtube "Born Free" yourself?
 
hhh
Suppose cylinder and Dirichet's conditions. Are $u(x,y)=\frac{Q}{4} \left(R^2-x^2-y^2\right)$ in $\mathbb R^2$ and $u(x,y,z)=\frac{Q}{6} \left(R^2-x^2-y^2-z^2\right)$ in $\mathbb R^3$ the only solutions?

D.C. are:

$$\begin{cases}
-\nabla u = \rho & \text{when A is an inner point} \\
u=0 & \text{when on the border of } \partial A \\
\end{cases}$$
 
@robjohn Can you please look at a pdf and comment on the appearance and TeXnical suggestions if any?
 
hhh
Reading page 809 here, bottom part where it only states the solutions without mentioning uniqueness.
 
@KannappanSampath sure
 
7:42 PM
@robjohn Ahhh... the memories...
 
@robjohn Here is a link: mediafire.com/?9tn6ob75gbbv85i Thank you for agreeing to look at it. : )
 
hhh
I cannot understand what it means $u=0$ on the border $\partial A$ when $u(x,y)=\frac{Q}{4}\left(R^2-x^2-y^2\right)$.
 
Lot of incomplete work, but you ca judge the style probably! @robjohn
 
hhh
@robjohn amazon.com/Born-Run-Hidden-Superathletes-Greatest/dp/0307266303 <--- for some reason, this book came to my mind from the song (running is often associated with youth, freedom -- etc). Wish someone had done some song about that :P
 
@tb Then this one isn't sane either... : )
 
hhh
7:51 PM
I am now not sure whether the Dirichet conditions are just premises...thinking.
 
@hhh You're kidding, right?
 
hhh
@robjohn No, I am not. The songs there do not have such melody/feeling as the earlier one...many have tried to do a song about running (related topics) but failed miserably...or I have not found any good yet.
 
Tiling window managers are so cool.
 
@hhh that you consider any good.
 
@robjohn My advisor told me how to do the homogeneous estimate :-)).
 
7:54 PM
@JonasTeuwen it's the grout that gets in my keyboard from them that annoys me ;-)
 
You just need to consider functions with mean $0$ as in the Poincaré inequality.
 
@JonasTeuwen how did he say to do it?
 
hhh
Could someone point me to the Wikipedia article about the Dirichet conditions above?
 
@robjohn Just consider functions with mean $0$.
 
hhh
en.wikipedia.org/wiki/Dirichlet_boundary_condition <--- this one does not seem to be it or?
 
7:55 PM
I haven't looked into it
 
@robjohn Into my .pdf you mean?
 
hhh
I cannot understand what my book is trying to say about this

$$\begin{cases} -\nabla u = \rho & \text{when A is an inner point} \\ u=0 & \text{when on the border of } \partial A \\ \end{cases}$$

so some English would help...
 
@hhh that seems to be what the DC means to me.
It is saying that $-\nabla u=\rho$ on the inside of of $A$ and and $u=0$ on the boundary of $A$.
 
hhh
@robjohn but what does it mean? They are just some assumptions but what about them? What happens if they are valid? Why are they called boundary value problem? I canot understand at all what is the goal here...
@robjohn but why? What is the purpose here? One defines such assumptions and then just states that the following $u(x,y,z)=\frac{Q}{6} \left(R^2-x^2-y^2-z^2\right)$ is the solution for cylinder -- smart but so what? Look how can I know that it is the only solution etc?
 
@hhh do they say $\nabla$ or $\Delta$?
 
hhh
7:59 PM
@robjohn $-\triangle u=Q$ is what is stated (p.809) and $\triangle=\nabla\cdot\nabla$
@robjohn This is not what the book is stating, it does not use nabla...
 
@hhh what is the context here? what is the purpose of making this statement? There is no way to answer your question without some context.
@hhh that makes more sense.
 
hhh
"the connection between field and source $\nabla \bar{F}=\rho$ can be written as $-\triangle u=\rho$" (top part 809), I fail here miserably to understand what the book is trying to say...it is like saying that $\nabla=\triangle$ which is of course non-sense.
 
@Kannappan the first definition seems a bit redundant. Is a linear operator on a vector space any less known than a linear transformation? Also, you write "$T:V\to V$ that maps $V$ to itself" which is also redundant.
Is it just me, or is the bookmark working better today?
 
Test: $x \sin x$.
Works better today for me as well, hence by $\text{strong}^{\text{strong}}$ induction principle, we conclude that book mark works well for everybody.
 
@robjohn "render MathJax" becomes un rendered sometimes... are you having that problem to Rob?
 
8:08 PM
It doesn't require me to rerun the bookmark :-)
not now
 
For me as well.
 
$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$
Whee!
 
@robjohn Maybe it's EMI again!!!
 
@robjohn So, how do I define it then.
The above thingy failed to render! : (
Induction doesn't tell you anything as $n \to \infty$.
 
@KannappanSampath what do you need to define?
 
8:12 PM
A linear operator on $V$ is a linear transformation from $V$ to itself, OK?
 
$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$
 
Failed to render even now. : (
 
renders and then becomes unrendered
ussually just after I send a message
it rerenders
 
It was working for me, and then I refreshed and re-ran the bookmark, now it is failing :-(
 
welcome to the club
 
8:15 PM
@Skullpatrol It has been doing that for a long time, but for the last bit it was working the way it used to.
 
@robjohn Indeed.
Periodic nonfuntionality?
 
@KannappanSampath Just saying that would be sufficient, I think.
 
OK. I'll edit that, Further comments, please.
 
Look, Brian is stuck in the basement with Asaf.
I shouldn't have listened to "Born Free" twice because now I can't get it out of my head :-(
 
@Kannappan: Lemma 2.1 seems to be missing some equivalent conditions.
 
8:24 PM
@robjohn Yes in fact a lot of places, it is in complete, I am sorry. I think I chose to give you at the wrong time. : (
 
@KannappanSampath perhaps I am looking at it too critically.
The general formatting looks fine
 
@MattN Hi
 
Hi.
 
What's up?
 
@robjohn Not at all. I'd like to know that, but given some are incomplete, I should have asked for it later as I observed. But how do complete proofs look. A few, I guess.
Asaf created a new Gallery room called " The Bin". I wonder why? (Likely, to transfer all my messages there?)
 
8:34 PM
What's the "topological countable product" of an ordinal?
 
@Matt Glad to tell you, I just broke 5k! :-)
 
Congratulations!
 
The recent answer has caught the fancy of many people, : P
 
@MattN The ordinal is equipped with the order topology and you take the product topology on that countable product.
 
8:37 PM
@tb Ahoy thar
 
Hi Skullpatrol
Hey N3 runs away from bear hug
@KannappanSampath congrats
 
@tb: you've shown up too early; I'm still here :-D
 
@tb Product topology of what? There is only one set, $\omega_1 + 1$.
 
Hi!
@tb Thank you! : )
 
8:39 PM
@MattN you take the countable product of $(\omega_1 + 1)$...
 
@tb the countable product?
How many times? $\omega$?
 
Let $X_1 = \omega_1 + 1$, $X_2 = X_1$, ... Then take $\prod_{n=1}^\infty X_n$.
@MattN doesn't matter. only the cardinality matters
 
@Skullpatrol Sorry about that: I went to lie down for a little and never made it back to the computer last night. No, I’m not familiar with Brown, Dolciani, et al.
 
@BrianMScott That's Ok.
 
@KannappanSampath In the proof of 3.6 you seem to be implicitly saying that $0^0=\operatorname{id}.$ You might want to state this convention explicitly.
 
8:43 PM
@ymar Hah! Nice, I did assume that. Thank you.
I'll add it in.
 
What an unproductive day. But at least I finally know why $z^n$. Or at least I think I do.
I still can't make sense of your hint though.
 
@MattN No, One's mind can make an heaven out of hell and hell out of heaven! It was a productive but not as productive as you wanted it to be. : )
 
I spent yesterday trying to learn complex numbers, and learn about the Gamma function.
 
: )
 
8:46 PM
Totally unrelated to anything to do with anything productive.
 
Heh. Disappeared again.
 
You forget: I’m on a dial-up connection. It would take me half an hour to listen. But yes, I do remember the song, though I don’t think that I ever saw the movie.
 
Is there any theorem or proof that if a number can not be written as a perfect square, then the root of the number is irrational ?
 
@BrianMScott I didn't know you were on a dial-up connection.
 
@Skullpatrol Ah, I thought that you were one of the folks around when it came up before. Never mind then.
 
8:48 PM
What do you think of this:
 
@BrianMScott Dial-Up!?
 
@N3buchadnezzar Yes.
@robjohn ’Fraid so. Rather good, as dial-up goes, but still ...
 
I’m going to have to switch this summer; too many sites are getting to be impossible.
 
@BrianMScott And it ties up your phone.
 
8:50 PM
@Skullpatrol That’s a feature, not a bug!
 
LOL
 
But it actually doesn’t, because I’ve two lines. That dates back to when I was married, so that we could both be on-line at once.
 
Do you have a cell phone?
 
when you were married ? Sad storry =(
 
As for the phone, I’m an unnatural creature: I don’t like ’em. And not only do I not have a cell phone, I’ve literally never used one.
 
8:52 PM
I don't like them either. : )
 
Amen.
 
63? Impressive
 
Wow, I thought I was the only one.
 
@N3buchadnezzar It was quite a while ago now. But no, it’s not much fun to be told two weeks after your tenth anniversary that your wife has decided that she’d like to try being single.
 
Our professor last year just turned 83, or 87 I can not quite remember.
@BrianMScott Indeed.
 
8:53 PM
@BrianMScott : /
 
@N3buchadnezzar It was a good time for me to retire, both in terms of benefits and in terms of the way education is changing.
 
Well better than being dumped for someone else, no?
@TeddyBär
Thanks!
 
@BrianMScott Education is changing?
 
@BrianMScott Do you think education is changing for the better or worse?
@Skullpatrol Ofcourse it is changing, it always is.
 
@MattN I’m not sure: being dumped for None of the Above is a bit disconcerting, too.
 
8:55 PM
@MattN why doesn't your $r$ depend on $x$ and $z$?
 
@BrianMScott Yes, I can see that. But nonetheless...
 
@BrianMScott Having to dump someone you still love is tough too.
 
@tb Yes.
 
?
 
@tb Ok. It's wrong. Thanks for looking at it : )
 
8:57 PM
@BrianMScott There are always plenty of more fish in the sea.
 
Totally realistic, skull.
 
I hate when people say that
 
@Skullpatrol At the college/university level it’s run increasingly as if it were a business with students as the product; this isn’t my notion of education. To me the good, small liberal arts college is as close to the ideal as an institution is likely to be able to get.
 
5151, @tb Shall we exchange our reps, numerologist? : )
 
@tb No actually. $z$ depends on $r$ and $x$. And $r$ is any old real.
@KannappanSampath : D
 

« first day (596 days earlier)      last day (4426 days later) »