Mathematics - 2011-12-16 (page 1 of 3)

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12:00 AM

Hmm... the current rate, in about two-three days I should hit 20k

Also, we've hit UTC midnight, so I am officially 500 days on the website.

Let me see.

visited 487 days, 418 consecutive.

500 days, 500 consecutives.

3

You're hardcore.

How do one find such statistics?

Obviously.

@N3buchadnezzar In one's profile.

I should hit the hay now, before I make a stupid decision like drinking more.

12:03 AM

That confirms my claim that you are one big cliche.

Night.

@JonasTeuwen You're an analyst, it's like halfway between a mathematician and a physicist/chemist/whatever. Your opinion doesn't count.

The more you drink, the better the idea of drinking more seems.

:(.

@N3buchadnezzar That is just not true.

@AsafKaragila Good night. And thanks.

12:05 AM

@Matt No problem.

Sometimes you think: "I will never drink again" when you drank enough.

@JonasTeuwen I think that every time I drink.

But that thought will never cross your mind while you drink.

Does it feel like your head is continuously falling?

And does the room spin?

@N3buchadnezzar Yes it will.

Hah. You people are weak minded and weak livered.

I have never uttered the words "I will never drink again." and I will never say them too. Or even think them.

12:11 AM

@AsafKaragila I think the $D_k$s are still broken.

@Matt Shouldn't be. Given a condition $(s,x)$ we have that $s\subseteq\max s$ and $\max s\setminus s = \emptyset\subseteq x$.

Perhaps the correct definition is: $$D_k = \{(n,x)\mid n>k\land x\subseteq\omega\setminus n\}$$

@Matt If this is not dense, then there should be a slight variation of this defining formula which would give a dense set. Either way, I am going to sleep... the alcohol is wearing off and I would not sleep without it.

@AsafKaragila Sorry was feeding the cats.

@AsafKaragila Good night!

I think I should go to sleep, too. Good night folks!

12:32 AM

Good night guys.

Good night @Matt.

user image

poundforce/inch^2

Only me that think that unit is sex related?

Wow... I barely posted an answer to this question, and the comments get updated to show that the OP doesn't want the solution that I posted (using mollifiers to construct a solution). As soon as I saw that I deleted the answer, but not before it gets downvoted. Almost every time I've been downvoted, a certain person is involved in the comments or answers. This person is very quick to downvote rather than comment.

12:52 AM

And they downvote without comment.

1:34 AM

Brave new algebra indeed...

Need more jurors to close this.

1 hour later…

2:48 AM

@MartinSleziak Thanks for this, Martin.

3:20 AM

Hello hello

Nice. Woke up and got two lhfs before even preparing coffee :)

Hi Srivatsan

Hey, hi tb

Hey, I have a question. Look at it?

Is the vector space of polynomials (seen as functions $[0, 1] \to \mathbb R$) closed in $(C([0,1]), \ell^p)$?

[closed in the sense of topology, in case that's not clear.]

Btw, I can show that it is closed for $p \geqslant 2$. I am not sure about $p \lt 2$.

Wow, a thousand views...

It can't be closed in any norm. In fact, a complete normed space has either finite dimension or its dimension is at least the cardinality of the continuum.

@tb digesting what you said...

What does the dimension of the space tell us?

3:35 AM

Maybe I misread you. What exactly is this $\ell^p$-norm?

[Well, may be I am not using the correct notations. Anyway..] So there is the space of all functions under the $\ell^p$ norm.

And the subset I care about is the set of degree- $\leqslant n$ polynomials.

Oops, I forgot about the degree in my first question; sorry about that. And I wrote $\ell^p$ when I meant $L^p$ :)

Well, that one is closed because finite-dimensional normed spaces are always complete, hence closed.

Um, ok. Thanks.

Oh well.

I can't make heads or tails of this.

The other thing I said can be seen (in a slightly weaker form) by a neat application of the Baire category theorem: If $E$ has countable dimension, consider $E_n = \operatorname{span}\{e_1,\ldots,e_n\}$. Then $E = \bigcup E_n$ and each $E_n$ is closed and nowhere dense. Hence $E$ cannot be complete, as it would be of first category in itself.

3:43 AM

He is somehow taking the derivative of a point on his manifold with respect to time.

Maybe this is some standard thing in physics that I don't know.

@Zhen Do you understand what he's after?

Doesn't make much sense to me either.

There is something interesting there though... is it possible to have a non-trivial fibre bundle $E \to B$, such that the trivial bundle $B \times U \to B$ factors through $E \to B$?

@Dylan: No, it makes no sense to me either.

Maybe he just needs to see an example of a non-trivial bundle over $TM$. Say, $TTM$.

On the other hand, maybe he's confused between a morphism of bundles vs the projection map of a bundle.

@Srivatsan Just to make sure: did you see my comment above?

(The other thing...)

@tb I don't remember which thing =)

That a normed space of countably infinite dimension can't be complete.

3:54 AM

Yes, I saw it. Is this an easy statement, or not?

Yes it is:

13 mins ago, by t.b.

The other thing I said can be seen (in a slightly weaker form) by a neat application of the Baire category theorem: If $E$ has countable dimension, consider $E_n = \operatorname{span}\{e_1,\ldots,e_n\}$. Then $E = \bigcup E_n$ and each $E_n$ is closed and nowhere dense. Hence $E$ cannot be complete, as it would be of first category in itself.

(or are you asking about the difficulty of Baire?)

@tb Oh, sorry, I didn't read this (I thought you were saying that to Dylan ;))

I see. That's neat, yes.

@robjohn yes, looks like it. I gave you a vote for the appreciation of the OP :)

@Srivatsan The slightly stronger assertion: a complete normed space of infinite dimension has at least dimension $2^{\aleph_0}$ is due to Mackey. A neat proof is here.

(behind paywall.)

Thanks for the link though, I will look at it when I am at my office.

Doesn't CMU allow off-campus access?

4:13 AM

@Srivatsan here it is in full:

user image

(It's nice when an entire paper can be quoted in chat... :D )

@JM I guess my favorite is this cutie

user image

@JM It does, but I haven't set up some stuff. =)

Wow, that Nelson one is really neat.

Thanks for both. :)

According to GEdgar and Halmos it maximizes importance/length

4:28 AM

@tb In that hilbert space projection question that you edited, there's one typo: "usin the Cauchy-Schwarz"

Chauchy -Schwarz :)

Yep, fixed both.

Is that the unholy offspring of Chaucer and Cauchy?

@tb Thanks.

@JM Is that the Canterbury guy?

The very same Old English writer. :)

4:32 AM

Not sure why there isn't more of a consensus here ...

I wonder why we need 6 answers to that one...

@TheChaz $+2i$ is the square root of $-4$ -- why?

"How to batter a dead horse..."

Convention. The square root is a function

The Chaz: Well, are you picking one of the branches of square root or something?

4:35 AM

Yes, the principal branch.

Chaz did say "principal" early on...

@JM Ok, just asking. =)

@JM batter a dead horse to make sure all the bacteria are dead, too?

It was just strange to me how the answers were... Well, like me talking about 1/x near x= 0!

@tb Apparently so.

4:36 AM

@JM Btw, how to decide the principal square roots of a general complex number , like $(3 - 4i)^2$?

Smallest positive angle?

Ok fine. Thanks.

That would give $-3 + 4i$ in this case, although I would have liked $3 - 4i$ to be the answer. Oh well =)

I like leslie townes's answers, but brevity and concision isn't their strong suit...

@Sri: Yep, Chaz got it. Generally you want the cut of your radicals to agree with your chosen cut of the logarithm.

@Srivatsan: didn't you give a neat argument for this one using AM-GM somewhere?

(is that a dupe btw.?)

4:41 AM

@tb I guess you are talking about this one: math.stackexchange.com/questions/64860/…

Exactly.

The one you linked to looks like this, but I am not fully sure. What's the k in that other question?

Just a positive integer, I guess

I think a bit of rearranging will give that. It's just the fact that $(1+\frac1n)^n$ is increasing.

@Srivatsan There's a "Let $k\in\mathbb N$ be fixed..." so t.b.'s right.

4:44 AM

Right. Thanks.

Well, it's more that $\left(1 + \frac1n\right)^{n+1}$ is decreasing. I don't like that all the answers are using $\log$. Looks like slight overkill to me.

I'm not entirely sure about it being a "dupe"; since some massaging is needed to turn one into the other...

"slight overkill" if there's ever been an oxymoronic expression...

No, I think it is $(1 + \frac1n)^n$ is increasing =)

5:21 AM

What's the largest possible volume of a taco, and how do I make one that big? :)

Hah.

$\mathbb{QUESTION}$

Neat question. :) You have a disk wrapped on a general cylinder...

(If this hasn't been studied previously, I'd be surprised.)

What is this trend of calling things "[insert subject] without tears?" Obviously, those book titles are lying

Yes, I think those titles are a bit of a swindle myself... :D

(Asking why the range of the inverse trigonometric functions are set that way would have made for a slightly more interesting question...)

Well, I found Topology w.t. rather pleasant to read [at least for an amateur like me]. =) So the book did what it advertised, I guess...

Not that I recommend the book though.

5:38 AM

I get 121 hits for without tears site:math.stackexchange.com...

Morris's book is certainly popular...

Are the 'tears' referenced in the title the type that drip from your eyes, or the type that rip things into pieces? =)

Esp. amusing when we're talking of topology...

If a book is bad enough, it is entirely possible that your eyes will tear up, and you'd feel compelled to tear up the book...

That was bound to happen... :/

5:56 AM

@Srivatsan I edited my question with a proof. I think it is wrong though.

@Potato Thanks, Potato. =)

I would very much appreciate if you could take a look at it.

@Potato I am looking at it. My idea was similar: bin the $n$ numbers into four bins: (real part positive, imaginary part positive), (real part negative, im. part positive), ..., (real part negative, im. part negative).

6:09 AM

@Srivatsan Do you see a flaw, or was the problem just poorly composed?

@Potato Well, right now, I am editing something else. Give me some time :)

6:38 AM

I see Lacey's paper was mentioned here. It's very nice a proof. I've mentioned this proof when talking about Hamel bases at our seminar.

These 3 questions are related to that topic: http://math.stackexchange.com/questions/74101/,
http://math.stackexchange.com/questions/33282/ and
http://math.stackexchange.com/questions/79184/

I learned about this stuff here at MSE.

@MartinSleziak Yes, this is very nice. If you like the idea, you might like Whitley's proof of Phillips's lemma ($c_0$ is not complemented in $\ell^\infty$). The observation that one can use the irrationals to produce an uncountable collection of almost disjoint subsets of a countable set is due to Sierpiński, I believe.

I am not too familiar with complemented subspaces. IIRC this means we it is direct summand and all spaces of finite dimension and cofinite dimensions are complemented.

Why is it useful to know whether a ssp is complemented?

Sorry @Potato. Was busy doing some other editing work. I will look at your question in some time (after grabbing something to eat).

@Srivatsan Thank you!

6:54 AM

To say that a closed subspace $Y$ of a topological vector space is complemented in $X$ is to say that there is another closed subspace such that $X = Y \oplus Z$. Equivalently, there exists a continuous projection of $X$ onto $Y$. Algebraically this is always possible, and in Hilbert spaces, too. A very deep theorem by Lindenstrauss and Tzafriri asserts the converse: If every closed subspace of a (real) Banach space $X$ is complemented then $X$ is linearly isomorphic to a Hilbert space.

ok, so I stand corrected - a direct summand where both spaces are closed

Is it correct to say, that if we know both $Y$ and $Z$ then we have simple characterization of all elements of $X$, which might be useful?

Yes, because if you don't require closedness, there's no restriction. Simply choose an algebraic complement (using a Hamel basis).

E.g. $c=c_0\oplus[e]$ is often useful. Although this one is very simple.

Yes. And this already shows a further intricacy: Even though $c_0$ is isomorphic to $c$, it isn't isometrically isomorphic to $c$.

$c_0\cong c$ you mean isomorphic as vector spaces?

7:00 AM

Isomorphic as Banach spaces.

There is a bounded linear bijection.

By isomorphic as Banach spaces I always understood isometrically isomorphic (=preserves norm).

It seems I was wrong.

There are two schools: I always mean the latter and add isometrically if I mean what you understood.

So if $X$ is a Banach space with norm $\| \cdot \|$, then $(X, \| \cdot \|)$ is isomorphic to $(X, 2 \| \cdot \|)$, but not isometrically? (I am clarifying the terminology.)

yes.

sorry afk

7:03 AM

An explicit isomorphism from $c$ to $c_0$ is given as follows if $(x_n) \in c$ let $x = \lim{x_n}$ and send $(x_n)$ to $(x, x_0 - x, x_1 - x, x_2 - x, \ldots)$.

I'm back

that was fast :)

I was thinking about that map.

It is obviously linear and bijective.

And $\lVert \phi(x) \rVert \le \lVert x\rVert$ so it's bounded.

No the last line was wrong.

It has norm 2

$\lVert \phi(x) \rVert \le 2\lVert x\rVert$

Ok, so it's isomorphism of Banach spaces in your terminology.

Yes.

So the second thing - to show that there is not isometry - is that relatively easy too?

7:07 AM

Did that question come up in MSE sometime back? Or am I thinking of something else?

Yes. You can see that the unit ball has no extremal points while the unit ball of $c$ has many.

Yes it came up about a month ago. I managed to forget what I just said :)

Srivatsan you're right it sounds familiar.

Ok, I believe I would be able to show this directly from the definition of extremal point.

@tb - what's an extremal point?

From what you (=t.b) wrote I thought that I should use $c \cong c_0 \oplus [e]$ somehow.

@Srivatsan I think the picture at wiki illustrates the definition nicely: en.wikipedia.org/wiki/Krein%E2%80%93Milman_theorem

user image

The points of a convex body which cannot be in the middle of some non-trivial line segment, with both ends in that body.

Aw, ok.

7:12 AM

In connection with Krein-Milman theorem (wiki link above) they can be often useful, to describe convex bodies in some sense.

here is what you're thinking of (sorry my browser hung)

And it is somehow related to Choquet theory, which Jonas mentioned a few times in chat. But I am not familiar with it.

@tb I obviously seen that answer (I see my upvotes on both question and answer). It seems that I forget things too quickly...

The idea of Choquet theory is pretty nice: In finite dimensions Carathéodory showed that every point inside a compact convex set lies inside some simplex spanned by extremal points. What about infinite dimensions?

BTW here's an unanswered question about complemented spaces: math.stackexchange.com/questions/86441/c-00-1-in-ck (I was not able to say much about it. Just in case it would interest you.)

Yes, I've seen it, and I've seen that OP found an answer to the question...

7:17 AM

@tb Why can't you just take a random line through that given point, and see where that meets the boundary of the set? Of course, it's not perhaps not that simple, but what am I missing?

@tb Yes, and in infinite dimensions instead of finite linear combinations we can use integrals with respect to probability measures to get convex hull IIRC.

what I know about Choquet theory/Choquet theorem is that it gives some kind of reformulation of Krein-Milman using these integrals

@Srivatsan You want the vertices of the simplex to be extremal points of the original convex set. Think of a cube for example. It is not quite trivial already in this situation.

@tb Oops, sorry. I don't know what I was thinking.

Which I believe can be sometimes useful, but I guess many things can be shown from the usual version of KM (using finite convex combinations and limits).

@MartinSleziak as you said, instead of finite convex combinations, let's take measures supported on the extremal points and look at their barycenters. Choquet's theorem says that this is possible, i.e., every point inside a compact convex metrizable set is the barycenter of some probability measure supported on the extremal points.

@MartinSleziak The salient ingredient is that you have a measure that represents your point. This isn't clear at all from Krein-Mil'man itself.

7:23 AM

@tb Yes, I see that it is more general. I am just saying that so far KM was enough for what I needed.

I first stumbled upon Choquet's theory when I saw an integral representation of some types of finitely additive measures in a paper. But later I found out the representation was wrong. But the authors did not refer to Choquet theory, they referred to some paper by Hewitt and Yosida.

@Potato, I went through the argument and could find nothing wrong. =)

Yes, the theory of finitely additive measures is cluttered with mistakes. Many people aren't quite aware of the many subtleties involved with them...

Anyway, thanks a lot for your time t.b.; I'll definitely have a look at the paper you mentioned when I have time. And - hopefully - I'll gain better understanding of complemented subspaces when studying it.

@Srivatsan Thank you!

Based on this quote, it seems that complemented subspaces are an important notion: The problems related to complemented subspaces are in the heart of the theory of Banach spaces and are more than fifty years old (Johnson and Lindenstrauss 2001). mathworld.wolfram.com/ComplementedSubspace.html

@tb Perhaps I will add some new mistakes :-) Anyway, the author of that paper was working in social choice theory, so he probably studied economics. I am not sure how high standards of mathematical knowledge is usual between economists.

But taking into account that Aliprantis was an economist and he wrote nice books on functional analysis, there are good mathematicians among economists.

7:34 AM

There definitely are...

7:45 AM

$\frac{1}{\pi}$? Who would imagine?

Yes, that's very neat. After that egregious non-answer that was posted first, this came as a nice surprise.

That's especially strange because this question has nothing to do with complex numbers really. I can think of this generalisation: suppose $z_i \in \mathbb R^d$ are $n$ vectors. Show that there exists $J \subseteq [n]$ such that $\| \sum \limits_{i \in J} z_i \| \geqslant c_d \sum_{i = 1}^n \| z_i \|$ for some $c_d$ independent of $n$. And of course, what does $c_d$ grow like?

@tb That one was hilarious. =)

I would've guessed that $c_d = 2^{-d}$, but that's down the drain now.

That's a nice question. Why not ask Robert in a comment?

@tb Yes, will ask.

8:00 AM

@Srivatsan thanks for that!

@tb Well, I should thank you. :)

Done.

8:13 AM

Don't they have like class policy and so on to decide such things?

Top of the not to you, and whatmorrow.

@AsafKaragila Hey Asaf.

What's up?

Nothing much. Some strange questions

Oh yeah, I saw that. I was like "Say whaaa?!"

8:20 AM

Good morning.

Off to work!

@JonasTeuwen Hi Jonas.

Bye Jonas

tb: Did you see robjohn's solution? What's up with this question? So many unexpected answers...

robjohn's answer starts off like Robert's, but it ends up with a different bound. =) I must confess I don't really follow either answer...

Despite being philosophically inclined myself, this question leaves a bad taste in my mouth. I'm thinking NARQ.

@Srivatsan well, I guess robjohn's solution is the one intended in the text but it is more wasteful than Robert's. (but still very nice)

8:31 AM

@tb Um, I think I can get $1/4\sqrt{2}$ exactly. I wonder if there will be interest in that result =)

sorry, robjohn gets that as well. But if he writes that part alone (ignoring the general $\theta$ stuff), the proof will be a bit shorter, I guess.

@ZhenLin Same here. I was thinking, well, maybe JDH sees it and has something nice to say, otherwise I fear the worst.

@tb Do you think a preemptive protection would be a good idea?

@AsafKaragila well, only a mod can do that. We'd have to wait 24h. However, I don't think it's necessary.

A good first step would be for the OP to tell us what he means by "the natural numbers"...

@tb Oh.

8:35 AM

See here: "The question must be at least a day old"

8:49 AM

@ZhenLin Good comment to the OP!

My hunch is that this guy is asking about the identity theorem for holomorphic functions. But the question is asked in such a confusing way that it's hard to tell...

@Srivatsan I got something a bit bigger than $\frac{1}{4\sqrt{2}}$, but nowhere close to $\frac{1}{\pi}$

@robjohn Yes, I was trying to guess the proof that the textbook had in mind.

@Srivatsan Robert Israel's linearization of the problem seems to do the trick.

@robjohn But a bit better than $1/2\pi$ :)

@tb Yes. However, to get a nice number $\theta=\pi/2$ gives $\frac{1}{4\sqrt{2}}$

8:54 AM

@robjohn what do you mean? It gives something massively better, right? I'm sure that the official solution is something simpler.

@Srivatsan My method maximizes $\dfrac{\theta}{2\pi}\cos(\theta/2)$

and the maximum of that is a bit above $0.1786$

@robjohn Er, it seems that we are talking about different things here. :-) But yes, I saw both the answers... =)

@Srivatsan Not massively: approximately $2/100$ better

$1/50$?

@tb I think Srivatsan is talking about RI's $\frac{1}{\pi}$

8:58 AM

@tb I think you should take the ratio, not difference. For instance if $d$ is large, then $2^{-d}$ is very different from $1.9^{-d}$... =)

Okay, then: 1.017 times better

Text book : OP : robjohn : Robert = 1 : 1.414 : 1.0103 : 1.801

??=$1.01032837641$

Ok, that number seems right. Sorry, I was confused because it is smaller than the OP's figure; I guess it should be... =)

However, I will take refuge in the fact that I was trying to show how the book came up with its constant; I was not trying to find the best constant.

Robert Israel's answer is along the same lines, but using linearity to get a better answer.

It's better than posting an answer, only to see that the OP has changed the question to specify that he doesn't want my type of answer, and being downvoted before I could delete the answer. :-p

9:09 AM

@robjohn Where did that happen?

Can you guys check if what I wrote here makes sense in the context of the question? math.stackexchange.com/a/91974/13425

Oh oh, tb was 23 minutes ahead of me, it seems =)

I am wondering if I am doing something circular...

@tb on this question

Yes. You are wondering if you're doing something circular.

@Srivatsan Looks good. I would mention that $B_n$ is closed by continuity of $f^{(n)}$.

@robjohn But your deleted answer shouldn't contribute to your rep, that is: recalculating should give you back the points. (and what happened with the votes to this question? They seem to be completely out of proportion...)

@tb Thanks, done that.

9:17 AM

@tb The votes do seem to be high, but it was not easy to get a solution. GEdgar got a very nice answer.

I didn't revisit the problem after the downvote (I believe it was Davide, since he downvotes quickly without comment, unless I query about the downvote).

@robjohn If he doesn't leave comments, how did you figure out?

You queried and he replied?

@Srivatsan If I query, he is the one that replies.

So I am assuming it was him.

@robjohn But Davide has only three downvotes...

@tb hmmm... he has always been there and commented on the downvotes I've gotten. I think I've only gotten 3 downvotes.

I find it hard to believe that he has only downvoted my posts.

@robjohn of course, only the posts that aren't deleted are counted there.

By the way: I'm not saying the question was easy and I like GEdgar's solution, but for a question that needed 7 comments to get to a clear question, 21 votes seem a bit much.

9:25 AM

@tb and I have deleted most, if not all, of the posts that were downvoted. I think I may have one left undeleted.

Usually, I see that my answer has been bogus and delete it. There was one or two that were good answers that were downvoted for some reason that eludes me.

@robjohn That they were posted in a public site could be the most satisfactory reason after all =)

@Srivatsan Perhaps :-)

@robjohn There are some people that use their votes for sorting the answers according to their tastes (I've seen that happening on MO). So if an answer that they like better has one vote less, they upvote it and downvote the other one.

@tb I am not surprised.

@tb How do you see such things? Can you answer in the abstract -- as Bill puts it?

9:31 AM

@Srivatsan Making assumptions based on upvote/downvote synchronization probably.

Um, ok.

I think there was a meta thread where some people said they did that and I distinctly remember one situation where I upvoted one answer among two that had some vote count and five minutes later their vote count was reversed.

See also Mike's comment here

Well, they should've closed that question. Anyway...

What are you talking about?

I think this question is better suited for Math Stack Exchange. It follows directly from looking at the generating series. – Eric Naslund Oct 19 at 19:08

9:37 AM

@MartinSleziak Yes, but that comment is not enough :)

Srivatsan: It was intended as an explanation for_What are you talking about_.

But I see that you included a link now.

Has there ever been anyone claiming that there is any sort of consensus what should be considered "research-level" or "suitable for MO"?

Maybe "quid" tries to achieve that. But if he continues this way he'll have written more than Google is able to search.

@tb But it's not like there was disagreement voiced in this particular thread. =)

@tb could there ever be an objective criterion?

@robjohn Yes. How about: It belongs where I say it belongs. =)

9:42 AM

@robjohn I don't think so. But opinions and attitudes vary very much from subject to subject.

(and depending on the OP)

we should start math.trashcompact.com for questions about division by zero, etc

3

But please include "easy FLT", Goldbach and odd perfect numbers there...

In other words, easily-understood speculative mathematics? :p

@tb sure, they are covered under "etc" :-)

The amusing thing is, there is probably some more esoteric speculative mathematics which would probably attract upvotes on MathOverflow.

9:49 AM

Definitely. There are some philosophical contributors to the n-lab/n-café that managed to do that already.

@tb I have a feeling you'll enjoy this comment...

hi

This is weird - the same person who answered a question, which looks close to a duplicate for me, asked the same thing: math.stackexchange.com/questions/91979 (There is a minor difference between the two questions, but still it's weird. Or I am missing something.)

@Srivatsan You mean that silly nitpicking? I'm all for silliness, yes, but that's a bit much.

hi VVV

guys would you believe this story?

10:01 AM

what story?

@MartinSleziak Ignore what I wrote, I was not reading carefully enough.

@MartinSleziak First the OP added the same constant to the numerator and denominator, now a different amount to the numerator and denominator. It's not a duplicate, I think.

two persons go swimming at night an old lady thinks they want to suicide calls the police, the persons are arrested, sent to the hospital and from there to the mental asylum.

@MartinSleziak too late for that. =)

oh and before they are sent to the hospital they are also imprisoned

10:02 AM

@Srivatsan BTW How do you link to your own message without pinging yourself? Is it possible?

@MartinSleziak No, we do not know of a way. Even linking to your own message is a bit of a pain. Not a single click thing.

"With Lophital I showed that" =)

hm... VVV that sounds quite urban legend-y to me.

BTW after deleting my (incorrect) comment about possible duplicate at math.stackexchange.com/questions/91979 the other question is still shown as linked. I wonder whether it stays that way, or it will disappear eventually.

@Srivatsan I wonder how many more uniform convergence questions Jozef has in his textbook.

@MartinSleziak Perhaps it will remain? Your comment is not truly deleted. It's all there somewhere... [In particular, the mods can see your deleted comments, I believe.]

10:22 AM

@Srivatsan: my answer may be ultimately the same as the integral answer by mjqxxxx, but I think it is cleaner.

@robjohn Nice. It has a flavor of both Mike's answer + mxpq's answer.

@robjohn [I cannot upvote you right now, btw.]

Again with the votes?!

You got 40 of them just 10 hours ago!!

@AsafKaragila votaholic. =)

@AsafKaragila Vote early, vote often... is a saying we have here on election day ;-)

@robjohn How does vote often fit for regular elections? =)

10:29 AM

@Srivatsan It's a joke... you are only allowed to vote once, but...

@robjohn It's early morning for me right now, so isn't it late for you?

@Srivatsan it is 2:39 here.

@Srivatsan 3 hours earlier than there.

@robjohn Well, I was wondering if you weren't going to sleep. Like me.

10:44 AM

Hi.

The optics professor is lightning fast in calculations! I'm writing a paper with him.

Hi again, Jonas.

Yes.

Optics professor? Is that like physics?

He stares for 5 seconds in the air, gives me a two line formula and says: this should be the solution. I calculate for two days and get that solution.

Yes.

@Srivatsan you are going to sleep just before the sun comes up?

10:48 AM

@robjohn No. A bit later. =)

I need to know what the standard reference for special functions is. And @JM is not here 8-).

Astonishingly, the message pointing to robjohn's ChatJax script has more stars than the answer on the link.

@AsafKaragila So more people like the link to the post than the post itself.

I guess.

@AsafKaragila or you more than me ;-)

10:56 AM

It is very reasonable! I'm lovable!

Nikhil Bellarykar

hi Mathematicians

I'm a lovable mathematician! Hooray!

Nikhil Bellarykar

I'm sure you are :D

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