Mathematics - 2012-01-26 (page 5 of 6)

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7:00 PM

@Srivatsan How many of them are there, how much change they caused (I mean Shelah completely rebuilt model theory, half of set theory, and probably some more stuff too) and how long before all the results will be understood. Erdos' work is pretty straightforward, Shelah's work will remain ciphered for years before people will truly understand what's going on there.

@AsafKaragila "Erdos work is straightforward." Not at all. Nobody understood the power of the probabilistic method quite like him, before him at least.

I should send an email to Arnie Miller. If anyone knows a better bound on nonmeasurable sets, it should be him.

@Srivatsan No one truly understands a lot of Shelah results after him as well. :-)

To compare the technical weight of Erdos' papers with those after him would be a bit anachronistic, I think.

Kannappan Sampath

@AsafKaragila As of now, Shelah has "only" 986 papers! :)

It's also very scary to talk to him. He remembers everything "Oh this? Paper 538, this one is 459 and that is 439"

@KannappanSampath There are final drafts of 1004 being prepared for printing or whatever. I saw those when I was in Jerusalem last Monday.

Kannappan Sampath

7:04 PM

@AsafKaragila I was just being......(what). I only wonder how people write so many papers!

@KannappanSampath People? It's just him.

How many Shelah's write so many papers? :-) (Answer: All of them.)

Sorry, I just deduced I am cranky. I'll be off for now.

Who in the world is this Shelah you all are talking about? He sounds like a fictional character.

Kannappan Sampath

@AsafKaragila True, But, generally writing more than say 12 papers is in itself achievement I think, Don't you agree, Asaf?

Depends on your perspective. Surely at first a paper at all is an achievement.

7:06 PM

@anon Try wiki: en.wikipedia.org/wiki/Saharon_Shelah

The really scary part is that he wrote the last 300 papers in the past decade or something like that. Most people would slow down, he picked up the pace.

@AsafKaragila For instance, Ken Kunen’s web pages list 106 papers and a review, and of course there’s his set theory book $-$ and he’s a fairly significant name in the field.

@BrianMScott Apter is at 200 or so; Truss at 100'ish; Jech wrote several books and a lot of papers too.

@AsafKaragila Out of curiosity: he wrote any books? I couldn't find anything off the wiki page, so perhaps the answer is no.

@Srivatsan A lot.

About forcing, cardinal arithmetics, model theory, and so on.

7:10 PM

@BrianMScott One review counts so much? You mean the review of a book, right?

How much time/effort does a review take? (Usually, from what you have experienced/heard)

Kannappan Sampath

Proper forcing seems to be a technique he introduced to the world?

Yes.

Have you written any reviews, Brian?

I mentioned it simply because it appears on the same web pages. It’s about five pages long, so it’s not a trivial piece of writing.

@Srivatsan No. I published a couple dozen papers and did some refereeing.

@Srivatsan shelah.logic.at/books.html

7:13 PM

@BrianMScott Just to clarify: I didn't imply any judgement on how trivial/nontrivial a review is. Asking to understand, is all... =)

@AsafKaragila Ah, now that is a lot of books, yes.

@Brian: You said you wrote a paper with a Shelah co-author. Who was that?

Kannappan Sampath

@Srivatsan Time taken to write a review is usually longer than that taken for writing the paper of same length. Largely because, a reviewer is expected to go through the book thoroughly pointing out what kind of course is the material in the book best suited for? What is expected of a instructor using that book? In what way is it different from extant books and so on...

@Srivatsan They're big books, too.

@AsafKaragila Rosłanowski. There was a third author as well, a colleague of mine at the time named Piotrowski.

Ah.

7:14 PM

@KannappanSampath "A review is longer than a paper of the same length." What.

:=)

@BrianMScott He's a good friend of Shelah, one of my friend doing his masters under Shelah and his thesis is related to their joint work about creatures forcing.

@KannappanSampath Yes, I agree with this also.

Creatures forcing?

The Pinched - Cube Topology

That’s the one. Pretty straightforward stuff.

7:16 PM

@BrianMScott I have no idea what that is really. I just know the name.

Do you know Miller in person, by the way?

@AsafKaragila It’s intermediate between the product and box topologies on a product. Arnie Miller? No, I left Madison before he arrived. Unfortunately.

@BrianMScott I meant I don't know what is creatures forcing... :-)

@AsafKaragila Oh, okay. Funny name, though.

@BrianMScott Shelah likes amusing names.

Shelah and Roslanowski has a joint paper "Sheva Sheva Sheva - Large creatures", where as sheva is Hebrew for 7, and 777 is the number of god (or something)

I can’t complain: I once wrote a paper about globs. (Generalized linearly ordered bases.)

7:23 PM

Is 666 like some incomplete form of 777?

Something like that.

What is six in Hebrew?

shisha, shesh

Oh, 7 makes sense. After all the world was created in seven days, hence seven days a week..

Shesh.

Shisha is when you say "six [masculine] things".

Yes, I had googled for it. That webpage gave some examples as well, but I didn't know which examples were for masculine and which one for feminine. =)

Hi, @Martin. I have to retract what I said the other day: I may have had that whole Herrlich exercise, but if so, there’s a part that I’m damned if I can reconstruct.

7:37 PM

Hi, Brian.

Did you mean Zbigniew Piotrowski?

I do indeed. Do you know him?

He was here in Bratislava several times. I believe he also has some joint works with Slovak mathematicians.

Which part, Brian? I'm sure it's more fun than the homework I have to submit a week ago.

Doing homework past the deadline date -- Not. Fun!

Yeah. Tell me about it.

It's algebra, no less!

7:40 PM

@AsafKaragila If $X$ is completely regular, and the canonical map $X\to[0,1]^{C(X,[0,1])}$ is closed, then every zero-ultrafilter in $X$ converges.

@BrianMScott I was at his talks at some seminars here. One was at a seminar led by Lubica Hola, another one at a seminar lead by Pavel Kostyrko. (But I don't think it's very probable you have heard these names before. L. Hola does research in hypertopologies and P. Kostyrko some real analysis, matrix summability methods and similar stuff.)

He talked about separate and joint continuity, quasicontinuity and similar things.

The canonical map is?

@MartinSleziak That’s Zbiszek, all right!

The canonical map is $x\mapsto \langle f(x) \rangle$.

Oh oh oh, right.

7:42 PM

@AsafKaragila $x\mapsto\langle f(x):f\in C(X,[0,1])\rangle$.

Perhaps $x\mapsto\langle f(x)\rangle_{f\in C}$.

Kannappan Sampath

@BrianMScott You mean Zbigniew?

I hope the notation looks standard enough.

@KannappanSampath Yes. Pet form of the name, though I may not have spelled it quite right.

@MartinSleziak Kostyrko’s name sounds vaguely familiar, though I’ve no idea why.

@BrianMScott He did lot of various things. I think I saw his paper cited in van Rooij-Schikhof for some result on symmetric derivatives or some related topic.

@BrianMScott To show that the canonical map is closed for compact Hausdorff $X$ all that I use is that continuous image of compact is compact and compact subset of Hausdorff is closed. No choice there, right?

By zero-ultrafilter you mean ultrafilter in the system of zero-sets, i.e. something closed under unions, subsets which are zero-sets and maximal, correct?

7:48 PM

@MartinSleziak But in that problem you don’t have a compact $X$: it’s only Tikhonov-compact. However, that’s enough to show that the canonical embedding is closed.

@MartinSleziak Closed under finite intersections and supersets that are zero-sets, and maximal with respect to those properties.

Of course, filter not ideal....

When you say zero set...

I noticed that you used the ideal in your notion of $\mathscr{I}$-continuous where I’d have used the corresponding filter.

Do you mean Borel?

@AsafKaragila A zero-set is the inverse image of $\{0\}$ under a continuous real-valued function.

7:53 PM

Ah.

@BrianMScott You seem to know more about me then I supposed. I believe I was continuing in work which was done by other people in a different paper, so I used their choice. It seems that both ideals and filters are used, perhaps with filters being more frequent. Nevertheless, I doubt there's something really interesting in that paper.

@MartinSleziak I just happened to look at your list of papers, and that one caught my eye.

The other one that caught my eye was the Fibonacci paper that you supervised. :-)

@BrianMScott It was my first time supervising B.Sc. thesis. (I don't thing I am very good as a supervisor.) Anyway, it's written in Slovak.

Which I don’t read, but I could follow enough of the mathematics to get a fairly good idea of what the thesis was like.

Supervising that thesis was reason enough for me to by the book about combinatorial proofs by Benjamin and Quinn.

BTW from Math.SE people, Mike Spivey has some paper on matrix proofs of identities about Fibonacci numbers.

8:03 PM

@MartinSleziak I’ve been tempted to get it; the bits and pieces that I’ve seen look good, and I’m very fond of combinatorial arguments.

@MartinSleziak I would never have guessed that his background was in optimization theory!

I like his hair. =)

Kannappan Sampath

@Srivatsan More than his Mathematics? =)

@BrianMScott Hm, he is one of the few people who answers linear programming questions here, at SE.

@KannappanSampath Well, yes, in fact. The photo is distracting. =)

@Srivatsan I hadn’t realized that, since I never look at linear programming if I can help it. :-)

@BrianMScott Hm, next time I run across a new linear programming question, I will force you to look at it then. =)

But, seriously: let me try to search that tag and see.

8:12 PM

anon has a nice answer.

Maybe I should have said: He is one of the few people I know of, who answers lin. programming questions.

Yes, slick.

But do you see what Didier meant? How is Cesaro applicable?

Meh.

@anon ? :)

Kannappan Sampath

@anon sweet answer!

@Srivatsan This, I think.

8:16 PM

Ah. Thanks!

Kannappan Sampath

I have a doubt: One of the privilege reads: Seeing Close Votes! I don't understand what this means at all! Can someone help!

[I think I'm fond of "!" these days! I never use "." at all!!.]

People with enough rep can vote to close questions. If a question gets five votes, it automatically closes. One privilege is being able to see how many close votes a question has at any given time.

Kannappan Sampath

@anon A question got closed now. I didn't and am not able to see any such thing!

If it's already closed you can even see the names of the people who closed the question under it. (And you don't need any reputation for that.) The privilege is useful before a question has been closed, where you can see the less-than-five votes cast.

Kannappan Sampath

8:52 PM

Can anyone shed light on this question? I really am confused with my basic group theory!

I made a new avatar for mma.se: gravatar.com/avatar/…

Gaudy!

@robjohn Wh.. What? What does it mean? :=)

@BrianMScott Opposite colors do that I guess.

@Srivatsan It is just a Möbius strip with a tube above one side :-)

Above one side? I think it's linked with the gray tube.

8:57 PM

@robjohn Ha, that's very cool. =)

@anon well, if it is above one side of a Möbius strip...

But why the glowing green?

@robjohn: D'oh.

@Srivatsan I am a freak; I like green and purple. But also mma.sse has a greenish theme.

@robjohn Oh, I did not know that (the greenish theme).

@robjohn But I totally agree with the first half of the first sentence. I mean, I can't agree more. =)

9:00 PM

The ping highlights are green and the round ping notifiers are green instead of orange as here

@Srivatsan :-p

@KannappanSampath I take it that you’ve now seen Arturo’s answer?

@BrianMScott to which question?

@BrianMScott never mind

Kannappan Sampath

@BrianMScott Brian, I saw his answer, But , I am feeling so dumb now!

(Gosh, what's up with me? I don't mean any of these things, but I do tend to overdo...)

Indeed, that wasn't very nice @Sri. You deleted that mighty fast ;)

9:04 PM

@anon Yeah. As I said, I don't mean it. It's just pulling the leg without any reason. I shall trim that bit from now on.

I've cut my hair.

@JonasTeuwen Hm. That sounds interesting.

Well, someone else did it, but that is a common way to phrase it in Dutch 8-): picasaweb.google.com/lh/photo/…

That is interesting. Why did you decide to get it cut?

picasaweb.google.com/lh/photo/…

Because I've had it quite long for over 10 years. So I was thinking, now that I have a job (PhD student is a job here) it is time for some change 8-).

9:08 PM

Just found out there's a "BMW algebra." I wonder what kind of mileage it gets.

2

@JonasTeuwen I thought you were posting a "before" and "after" photo, but no, it wasn't that.

I have plenty "before" pictures.

Why why whyyyyyyyyyyyyyyyyyyyyyy why did you cut your hair?!!?

Because of ^.

I like the title!

9:09 PM

I should add this as a channel rule: No trimming one's hair!!

I'll take a beard instead.

*Facial hair.

You should know that the aggregate length of all hairs on one's body is not a constant. If you cut your hair, it won't grow your beard out.

@AsafKaragila Aha, in that case, I would've been banned like thrice from this room.

@Srivatsan You are a Hindu, I guess I can live with that.

9:11 PM

He should have been a Sikh.

Now that's sick [sic]. ;-)

@AsafKaragila More than that, I am my parents' son. I was literally forced to visit the barber first thing I landed home in summer. Notwithstanding the jet lag and all that.

@JonasTeuwen But then he’d have to Singh for his supper.

@Srivatsan I have not met someone who was not his parents' child.

Kannappan Sampath

@Asaf That's hard to have met someone. But, I have a proof that justifies existence.

9:15 PM

How many people have you met that are Sri's parents' son?

@anon (raising the hands) I know, I know.

Any guesses? You can pick a random number from 1 to 4 if you like.

@KannappanSampath Does it use the axiom of choice? ’Cause Sri apparently had no choice.

Kannappan Sampath

Your bro/sister or ...

@Sri

Jacopo Notarstefano

Sri.siblings.count + 1

(It might be off-by-one, I'm not sure about the meaning of "siblings"...)

Kannappan Sampath

@BrianMScott Yes, It has to use Axiom of choice apparently :(

9:17 PM

@KannappanSampath Hm, sorry, I didn't follow this..

@JacopoNotarstefano I don't think one would be considered one's own sibling.

Kannappan Sampath

@Srivatsan I wanted to say that the number of such people satisfying Anon's quest is the cardinality of the set of your siblings. Typed out such a short sentence in haste!

Jacopo Notarstefano

Wiki gives the definition "shares at least one parent with", which clearly wasn't written by a mathematician : )

@KannappanSampath Ah, I was asking for guesses on that number. It seems nobody has any.

Anon's quest Oh boy do I like the sound of that phrase!

Jacopo Notarstefano

9:20 PM

I'd go with 4, because that's how many I have!

Kannappan Sampath

I think it would be 1!

@Srivatsan]

@Srivatsan I'll guess 2.

@KannappanSampath You are guessing that I have a sibling or that I am a lone child? Apparently there is some confusion about what number we are guessing.

(I have a brother, not sure who is right and who isn't.)

If Srivatsan has siblings as much as he has deities then he has infinitely many of them :-)

3

Kannappan Sampath

@Srivatsan I guess that you and your sibling are the only people!

9:21 PM

@Srivatsan I am guessing 2 siblings (Sri + 2)

@robjohn Hm, one off. K. is correct. :=)

is it Sri + 1?

Kannappan Sampath

@AsafKaragila That seems like a one line definition of Hindhuism?

@AsafKaragila Ha. At least we are not counting the number of hands on the said deities. =)

@Srivatsan Would that be uncountably infinite?

Kannappan Sampath

9:24 PM

@Srivatsan But that shouldn't make the problem harder! Given, arbitrary union of countable sets is still countable!

@AsafKaragila I could never count them. :D

An arbitrary union of countable sets needn’t be countable, even with choice!

I actually believe that the universe is an amorphous set. It's not finite, but you can only talk about finite pieces (or their complement) of it.

How about all the singletons in $\mathbf R$? 8-).

Ah, of course, since every set is a union of singletons, aren't they?

Jacopo Notarstefano

9:25 PM

Countable union of countable sets is countable. Do we require a form of choice to assert this?

Of course.

Yes.

One needs to choose the enumerations of each set. I think that I wrote about that at least twice on the main site.

My collection of siblings occasionally felt countably infinite. (I’ve three brothers and three sisters, all younger.)

Hmm, well, what if the sets are subsets of the positive real numbers?

9:27 PM

Doesn’t help.

@JacopoNotarstefano I indeed wrote about that here.

Few months back (while studying the Vitali set construction), I had this question: how much choice do we need to assert that the vector space of the reals over the rationals has a Hamel basis?

@BrianMScott No? Can't we pick the one with the smallest element?

@Srivatsan Quite some choice.

Jacopo Notarstefano

@AsafKaragila Thank you!

9:28 PM

@JonasTeuwen If the continuum is a countable union of countable sets, then of course it's not enough.

It is true that the countable union of finite subsets of $\mathbb R$ is countable.

Because then you can use the linear ordering of $\mathbb{R}$.

Indeed.

@AsafKaragila Of course, I dare not post this question because I don't think I can follow the answers. But I thought the question was an interesting one. Do you think this could've (or, would've) been studied before?

@JonasTeuwen What’s the smallest element of $\{2^{-n}:n\in\omega\}$?

@Srivatsan I'll look it up and see if anything pops out.

9:31 PM

@AsafKaragila Do you have easy access to Rubin & Wossname?

@BrianMScott Right, I was thinking about that.

@BrianMScott Equivalents or Consequences?

@AsafKaragila Thanks for this, I should add that you don't spend too much effort into that question. It was just a random one.

Are there two now?

Was that meant to be Howard and Rubin?

I am not sure now - is one of the two books Rubin & Rubin...?

9:33 PM

Yes, I was thinking of Howard & Rubin, though I’ve never actually seen it. I did browse through Herman & Jean years ago.

Rubin, Rubin: Equivalents of the Axiom of Choice; Howard, Rubin: Consequences of the Axiom of Choice

There are two editions of Equivalents of the Axiom of Choice; and there is one more Consequences of the Axiom of Choice.

Hamel basis for R over Q is form 367 in H-R.

@Srivatsan In Herrlich's book there is a diagram in which it shows that $\mathbf{AC}(\mathbb R)$ (choice functions from subsets of $\mathbb R$) implies Hamel basis exists.

This is of course not surprising, it is also equivalent to $\mathbb R$ being well orderable - which in turn makes an easy proof for finding this basis.

Yes, I agree with it.

@MartinSleziak What does this mean? That Hamel basis for R over Q is equivalent to choice?

9:39 PM

No, it follows from some choice.

No, because $\mathbf{AC}(\mathbb R)$ is weaker than full choice.

@Srivatsan Howard-Rubin is a (big) book where many consequences of AC are collected, together with plenty of references and list of models, in which some of the forms hold/don't hold.

I'm not sure that the assertion that it exists is even equivalent to $\mathbf{AC}(\mathbb R)$, probably the Hamel basis existence is weaker.

@MartinSleziak Oh, this is the "consequences" book then, not "equivalences". I was confused that your comment disagreed with Asaf's (the in the next line).

@AsafKaragila Ah, Ok. Thanks, all three of you. Appreciated.

@Srivatsan Interestingly enough, your question made me understand a bit more when non-measurable sets can occur.

9:51 PM

@AsafKaragila As in, under which choice assumptions can nonmeasurable sets occur?

@Srivatsan Yes.

Cool. Glad to be of some use. =)

Asaf: You mean Existence of Hamel basis $\Rightarrow$ existence of non-measurable function (as a solution of Cauchy equation) $\Rightarrow$ existence of non-measurable set?

Jacopo Notarstefano

Does anybody know how does the inverse Ackermann function arise in the analysis of algorithms?

@MartinSleziak Exactamundo.

9:58 PM

My life is nothing without AC.

What happens when you'll die and find out there was never any free will, and thus you had no real choice all your life?

@JonasTeuwen You're not married, are you?

@Srivatsan: no other bites on your question

Bite? What does that mean?

@robjohn No, not yet.

I need a drink.

@Srivatsan No one else has answered it.

10:11 PM

@robjohn Did you just made Jonas sad by asking that question? :=)

You’re fishing for answers, and no fish has yet bitten (taken the bait).

@JonasTeuwen I should make some tea.

@Sri: Have you ever been fishing?

Hmm, I just made some coffee :-). FIlter Kenia The'Ri (AB PLUS).

Now Glendronach.

Send me some of that Glen, jerkwad.

10:12 PM

@JonasTeuwen that doesn't sound like coffee...

Yeah, that's Glen. Whisky.

Maybe "coffee" is the new euphemism kids are using today...

@AsafKaragila I have a whole collection of Glen's.

No, Kenia The'Ri is a coffee bean.

@anon No. (But I would've understood the reference under "usual" circumstances. Now sure why I didn't get it now. I think I use the idiom slightly differently; not sure how.)

Oh, I thought you say "I'm having coffee" when you actually mean "I'm chugging Glen in gulps"... :D

10:14 PM

@JM Yes. "The drinks are excellent at this coffee shop here." (At least it would fool me, I think.)

@JM Well, I'll stand by my tea. Perhaps jasmine green tea.

@robjohn Yes, but you did give a complete answer.

BTW, why don't you at least mention an asymptotic approximation for that expression? @rob

@Srivatsan Perhaps there isn't another way of getting the answer.

@robjohn Well, let's see. In case, it catches David's attention.

;)

@Srivatsan easy... $\dfrac{1}{2\sqrt{\pi d}}$

10:17 PM

@robjohn Yes, you could mention it in the post and compare it to the $\frac{1}{2d}$ lower bound mentioned by the OP.

Hang on! Wait a minute!!

What happened the the mean square?! Are you suddenly nice?!

@robjohn Wait: doesn't this depend on the parity of d? Do we get such a uniform asymptotic for both odd and even values?

@JM: Your avatar looks cool. It looks like a snake from a distance, but zooming in gives the intended effect.

Jacopo Notarstefano

Wow! That's a surprising result.

@robjohn Is that you Rob, or did someone take the square root of the "mean square" to find the root of all evil ??? ;-)

@Srivatsan Hah, you noticed the scales? I liked it... :)

Kannappan Sampath

10:22 PM

How do I center an image in the main site?

1000 views for this? People are gruesome... :P :D

@Skullpatrol Pardon? are you talking about on mma.se?

A bigger version of my Gravatar:

user image

2

@Kannapan: You can't, as far as I am aware. Put $\hskip 2in$ to skip two inches in front of an image. (Vary "2" as necessary.)

@JM very snakelike

10:24 PM

Looks a bit like dog dung, frankly ;)

3

Are the hexagons flat, or are they a bit concave?

Kannappan Sampath

LOL@anon for your dog dung comment!

@anon It's not that twirly, you know... :D

A hose made by bees?

@anon do you have the dog who lays the golden turds?

10:25 PM

@robjohn concave.

@JM they appeared that way.

Jacopo Notarstefano

What's the shape of snake scales, usually?

@BrianMScott Yeah, I was going for honey/amber coloring.

@robjohn: Yes. He sleeps next to the goose that lays golden eggs. Also, does one really need to censor "turds"? Heh.

@BrianMScott It will start squirting honey when you open the tap. =)

10:26 PM

@robjohn Heck, I'd be way interested in a dog whose turds are always translucent...

@anon we'll see :-)

@JM I’d say that you succeeded.

Bobert, why did you change your parent user to Mathematicae.SE?

I'm wondering if this is the best of JM's avatar till now. But I remember only the Cheelix, so I would give the award to this without second thought. =)

Who? Me? Probably not.

10:29 PM

@anon Anyway, a good thing you brought it up; I never would have realized that my previous avatar looked like mac and cheese myself... :D

@Srivatsan Challenge? ;)

@JM Who are you challenging? The competition is between your gravatars... :)

Jacopo Notarstefano

robjohn's "Mean square" is hard to beat, in my opinion.

@Srivatsan It's like you want me to outdo myself... :)

@JM Ha, I thought I called it "cheese".

@Srivatsan I'm confused; it was either you, anon, or Matt...

10:31 PM

I don't remember anymore. Let's just split the dish three ways.

Oct 23 '11 at 13:51, by t.b.

@robjohn: you're mean squared :p

Don't be so two-dimensional, Rob. =)

user image

Does anyone else find the new Google Plus logo too bright? Brown is a little distracting.

I'm not the one who called it cheese. Also, why is g+ brown? Why not green or red or something?

Jacopo Notarstefano

brown? Check your monitor. : D

@JacopoNotarstefano What is it? Maroon?

10:38 PM

What's it supposed to be? Looks brown on my lappy.

That looks like one of those shades nobody's sure with the name...

Well, I am not talking about the subtle question of which shade it is. I just find the colour a bit too bright. Anyone agrees?

@robjohn Before you were mean-squared and now you are not-squared ... but I do like the new look ...

Jacopo Notarstefano

#BE5F40

Well, it doesn't look like one of those spasm-inducing colors to me; I'm not uncomfy with it...

10:40 PM

Yes, I think the icon stands out way too much. Very annoying.

(I'm not fond of having too many social site linkery crap being festooned on questions, but that's a different issue.)

@JM I can tolerate it. But it distracts attention from the question, is what I feel.

Jacopo Notarstefano

I'm convincend that when searching for an hex code Google should return the color.

@JM Sadly, that might not be a reversible thing.

Try googling the phrase do a barrel roll

10:42 PM

...is it a new Google bomb?

Jacopo Notarstefano

Nope. A google easter egg.

Sometimes I think the Google folk have way too much time on their hands... :)

Jacopo Notarstefano

(Available on modern browsers, I suppose)

@anon I almost fell off the chair ...

nostalgia'd

10:45 PM

I still think their best fun stuff were the Pacman and Les Paul Google doodles...

@Skullpatrol I am still mean-squared here. I am just half-twisted with one side tied on mma.se

Google has fun April 1st features.

Does SE have any Apr 1 fun things?

@JM Our analysis group has three special function guys! :-).

@JonasTeuwen In general, or do they specialize in certain classes?

@anon I think they confine the nutty stuff to meta.SO...

Oh, I don't know that much detail, I can ask them tomorrow. I have just found out. The other guy in my office has a PhD in special functions but he didn't mention which ones.

10:49 PM

One time, they replaced the generic Gravatars with unicorns and ponies.

Jacopo Notarstefano

I remember something about ponies.

@JonasTeuwen Koornwinder's retired, IIRC, but still working on stuff. Is he one of those guys you talked to?

@robjohn Are you a half-twisted mean-square or a mean-square half-twisted?

@JM No, but the one in my office has a PhD with a guy that has a PhD with Koornwinder 8-). (He got his PhD at the same university)

@Srivatsan At least for beta sites I think I can safely say they use maroon for the Google+ link...

10:52 PM

@Skullpatrol Hmm, now I need to put my mean-square gravatar on a möbius strip...

@JonasTeuwen Twice removed. Close enough. :)

It is left-right symmetric, so it should work.

@robjohn Texture[] would be your function to use. :)

@JM The mathematics genealogy project tells me that they are all either children or grandchildren of Koornwinder!

Kannappan Sampath

Is this answer of mine good enough?

Comments welcome!

10:54 PM

@JonasTeuwen I'm not surprised... he's basically THE Dutch guy in special functions...

@robjohn How about forming a Klien bottle out of the mean-square?

Kannappan Sampath

@Skullpatrol Klein trembled in his grave!

@Srivatsan : I have updated the answer.

@KannappanSampath I'm trembling just waiting for his answer.

Kannappan Sampath

:)

10:56 PM

@robjohn Thanks. I wll take a look at it.

In mathematics, the Klein bottle () is a non-orientable surface, informally, a surface (a two-dimensional manifold) in which notions of left and right cannot be consistently defined. Other related non-orientable objects include the Möbius strip and the real projective plane. Whereas a Möbius strip is a surface with boundary, a Klein bottle has no boundary. (For comparison, a sphere is an orientable surface with no boundary.) The Klein bottle was first described in 1882 by the German mathematician Felix Klein. It may have been originally named the Kleinsche Fläche ("Klein surface") and th...

My son sent this:

user image

I posted in meta: about the colour of the G+ logo.

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