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1:27 PM
I hope everyone's weekend is going well. I had a long day yesterday. I took care of a lot of yard work, my son flew in from Chicago for the summer, and we had dinner when we got back from the airport. I was working on a question that got solved in pretty much the same way I was looking at it, so I needed to solve it another way. I need to stop working on questions with long answers. They take too long and usually get few votes.
@J.M.: how are things on Mma? I should pop over and look at what sorts of questions are being posted.
 
@robjohn Pretty good, I think. The activity isn't as frenetic as on math.SE, so it's much easier for me to catch up on questions I missed when I was away...
@robjohn I guess it depends on the topic. For the really specialized topics, the effort for writing is often in inverse proportion to the number of votes it gets. I've accepted it now, but it took me a long time to do so.
 
@J.M. Oh, I will not change. The questions with the long answers are much more rewarding personally.
 
(That's why I'm conscientious enough to vote questions/answers on specialized stuff that I understand. For instance, I don't think there is a question that I haven't voted on (not counting mine, of course).)
@robjohn "much more rewarding personally." - and that's why I'm picky with questions myself. :D
 
@J.M. Yes, that cuts down the number of questions answered, but it makes life more interesting.
 
Yeah, LHF is nice, but it can get dreary quickly...
 
1:39 PM
Any asymptotic interesting and elementary?
 
There is a very disappointing side effect of math.stackexchange
In that the answer to so many questions are available here.
Even some of these
 
@RagibZaman (If you know how to search, that is...)
 
[This] Math.se thread completely answers question 1 of that competition.
This , and Yes I do =] !
 
(You know about being able to press the up-arrow key to edit stuff, right?)
 
@BillDubuque Hi Bill. Last time you saved me when I thought that prop. 3.11 in Atiyah-MacDonald was boring. Now I'm reading the chapter about composition series of modules. Can you save me again? I am not enjoying this chapter at all.
 
1:45 PM
Is there a way of hiding a thread for 2 weeks until the competition ends?
 
@RagibZaman You could delete and then undelete later, but that's only for your stuff.
 
Moderators can temporarily delete the thread. Or if you downvote it enough it will disappear from the main page...
 
On the other hand, the bit Ragib linked to was last touched in February, so it has been off the front page for quite a bit...
 
Do you think this case warrants temporary deletion? I sure hope so, as the number of people who stumble across that page in the next 2 weeks and are entering the competition probably far outweigh those not entering...
 
@RagibZaman I think that's one of those times to ask at meta...
 
1:48 PM
One of my friends also entering to competition just sent me that link and asked me if I knew about it, so I fear many others know about it already.
 
(On another note, couldn't they have just come up with something completely different?)
 
You mean my friend? He was searching online for related problems for ideas and found that today.
 
@MattN. I have seen the Hilbert Basis Theorem and its proof :D
 
@RagibZaman Oh, I thought you had something to do with the contest organizers...
 
@RagibZaman You should tell henderson about such threads
I notify my lecturer of threads on se
 
user19161
1:54 PM
@FrankScience I was being deliberately vague to add some mystery.
 
@J.M. No, I'm upset because I'm entering the competition and worked very hard to solve that problem.
 
@RagibZaman Do not let external circumstances control your happiness
 
@RagibZaman Ah, an entirely different kettle of fish...
 
user19161
@robjohn Oh dear. Well, since you have so much rep already, you may let others have a chance to earn some rep sometimes!
 
@MattN. Actually I have found that leaving some stuff for a while and coming back to it say one year later or something can be very refreshing
 
1:57 PM
@JasperLoy I wouldn't say that I have taken anyone's opportunity for rep. Yesterday I was contending with joriki for a question, but he has 59k, so I don't worry about him :-)
 
@JasperLoy You did not mean basketball/football or something like this. Am I right?
 
user19161
@robjohn I see you are quite serious about earning rep here. It is indeed an exciting game!
 
user19161
@FrankScience Well, the humour comes from the possible naughty interpretation and the inherent ambiguity. As always, what I say in this chat may be interpreted in over 9000 ways.
 
@JasperLoy If I were serious about it, I wouldn't be answering the questions I do. Once in a while I take a breather and answer some lhf for a while, but most of my answers get only a small number of votes.
@JasperLoy and that is why we call you BabeLoy :-)
 
user19161
@robjohn Hmm, I think those that get the most votes are usually simple but not too simple questions.
 
2:01 PM
@JasperLoy Yes, that legendary "comfortable middle"...
 
user19161
@robjohn Before the edit, I was wondering which interpretation of Baba to apply, since Wikipedia gives me too many choices.
 
That is undoubtedly true, look at the most upvoted answers for top 50 rep member and it is almost always a simple question.
 
@RagibZaman what is?
 
user19161
@J.M. Yes, too trivial and people think it is no big deal, too hard and people don't know the stuff or don't bother to read.
 
@BenjaLim Jasper said "I think those that get the most votes are usually simple but not too simple questions."
 
2:03 PM
exactly.
Like I don't get why I get 45 for like arithmetic progressions
 
user19161
What happens here is just common sense. Psychologists take simple phenomena, apply technical terms to them and make it sound as if they know a lot about the human mind.
 
@RagibZaman and i see you're even more pro with the $x^3 + x$ thingy
 
lol I was hoping to double the accepted answer and get that gold badge!
 
user19161
But actually we just need to observe and think about things a bit and we understand why many things happen the way they do.
 
@JasperLoy well, I wasn't sure how to go with it. I wanted it to sound like Babylon but "Baby" would overshadow any intended meaning, but the main idea was the Tower of Babel. I wasn't sure if the "Babe" would have the same problem that "Baby" would. So I vacillated.
 
2:08 PM
@BenjaLim I don't suppose you're wondering why your answer on the sum of the odd numbers is so highly-rated? ;)
 
yes
 
I knew Eric's answer in that thread would become highly rated if he added that picture, but not that highly rated...
 
hmmm
well
that is why math.se voting patterns are so weird
 
@BenjaLim ergo, don't bother rationalizing them; you'll just drive yourself bonkers.
2
 
haha
 
 
2 hours later…
4:22 PM
Wow, I feel wasted man! 8-).
Does anybody know how I could test if \csname\currfilebase\Introduction\endcsname is empty?
 
@JonasTeuwen Sorry - not a TeX guru here :(
 
@OldJohn No problem! But a gardening guru? :-).
 
\ifstrempty ?
 
@JonasTeuwen I have a bit of success there :)
 
@ZhenLin It is not really empty it apparently can be \relax...
 
ami
4:27 PM
Hi
How can I find the coefficient of $[x^ny^m]$ where $1 \leq m \leq n$
$\frac{\log(1/(1-xy))}{(1-x)(1-y)(1-xy)}$
 
@ami Please ask on the main site, it is too long of a question.
 
ami
@JonasTeuwen ok
 
ami
4:45 PM
@an
@anon You there?
 
I just answered on the mainsite
(yes)
 
ami
@anon I just have modified the question a bit. Added one more series.
:)
 
neither equation seems to correspond to one in the paper, though they come close
 
ami
look at page no. 3
F(z,u)
left side
bottom-left *
 
I see $F$ on the bottom-left, and I stand by my comment
it looks somewhat like the first two terms, but the first two terms have numerators
 
ami
4:50 PM
yes. they are not same.
yes, they come close.
 
that's intentional? just want to make sure.
 
ami
yes.
$$ \frac{\log(\frac{1}{1-x})}{(1-x)(1-y)(1-xy)}$$ --> its nasty
could not able to solve it.
 
we'll see about that
@ami solved the second
 
ami
@anon Excellent. Although I could not understand the Sum's limit from n-m to n
Can you please elaborate it a bit
 
$n-k\le m\implies n\le m+k\implies n-m\le k$
 
5:20 PM
Hi there.
@Gigili: why the sad avatar? What's up?
 
@MattN. I told her it was gloomy.
 
@PeterTamaroff But what happened?
 
@MattN. No idea.
 
5:53 PM
Anyone here is against animal testing?
 
@PeterTamaroff Why do you ask?
 
@PeterTamaroff What a question. Of course. I mean: we already know how to make lip stick, we don't need to test it any more.
Also: the food I feed my pets is good enough. No more cats need to die to marginally "improve" on it.
Man, picture this: you're locked up in a tiny cage all day, sometimes you are starved, and daily you're probed and prodded. Until you're put down.
 
Heh, "improve". Though, how would you know if they'll eat it if you don't try to feed it to them? ;)
@MattN. A sucky life, that.
 
As for meds: I guess it's ok to feed pills to rats and see if they die and then only test it on hoomins if they don't.
@J.M. Well. I think you understand what I was saying. One manufacturer (owns the brands Iams and Eukanuba, cat & dog food) is famous for animal testing. Seriously, what pet owner can buy food for which pets have been tortured?
 
@MattN. Ah, yes, I know of those. Inexcusable.
2
If they won't eat it, then maybe because the blasted thing is unpalatable? Ugh.
 
6:05 PM
The sad thing is, I bet the majority of pet owners doesn't even know that they do that.
 
@MattN. LOL
 
Now cosmetics, on the other hand... maybe the PETA people will be happy if we test the cosmetics on the models they use for their ads.
 
Starred. Inexcusable is the word.
@J.M. The Body Shop sells tons of cosmetics that's been made without animal testing.
 
@MattN. I'm only for animal testing when it comes to curing disease and stuff.
 
@PeterTamaroff I'm not. There are too many people on this planet already. The more die, the better.
Humans are despicable.
 
6:07 PM
@MattN. Spoken like a true misanthrophist!
 
@PeterTamaroff Well. I don't a priori hate them. I just really despise them. They disgust me. Truly.
 
@MattN. And you happen to be some sort of alien lifeform?
 
@MattN. Agreed - I like the suggestion of taking all warning labels off everything and let nature take its course :)
 
@PeterTamaroff Nope.
@OldJohn : )
 
@MattN. I just think it is true the world is full of people... but they can have a good life if we weren't despicable.
 
6:10 PM
@MattN. sort of "survival of the smartest"
 
@OldJohn Yes, I understood. I assume you know Darwin awards : )
Anyway, gotta go eat dinner : )
See you later.
 
@OldJohn Deliciously devious. That "sue for the slightest injury" thing is totally getting out of hand...
 
@J.M. Hehehehe
@J.M. I think you ahve this in your profile, don't you?
 
user19161
@PeterTamaroff Sounds like something you wrote Pedro!
 
@PeterTamaroff Yes, I do. Nice, no?
 
6:20 PM
@J.M. It is quite a fun thing to read.-
@JasperLoy I'm pretty tolerant, a$@#ole
 
user19161
@PeterTamaroff At first I thought it was some silly article but then I liked it!
 
user19161
I just realized the Olympics is starting in a few days!
 
@JasperLoy The torch is being delivered just now!
 
user19161
@PeterTamaroff Yes, by Piotr the Great.
 
6:36 PM
@JasperLoy Great man, great man.
 
6:48 PM
:5463682 Hey Bill ?¿
 
@PeterTamaroff I was searching for a link to the post of Eric that someone mentioned got upvotes due to an image being added. Do you know it? Aargh this browser is fried, can't see what I'm doing. Hold on a sec. Ok, I'm back.
 
@BillDubuque On Main?
 
@PeterTamaroff I'm not sure. JM wrote above "I knew Eric's answer in that thread would become highly rated if he added that picture, but not that highly rated..."
 
@BillDubuque Oh, that. I was referring to this.
 
@J.M. Thanks. I thought it might refer to something recent I missed.
 
7:03 PM
@J.M. Nice TeXing!
 
@PeterTamaroff The kudos should be going to Eric... ;) I just suggested that he add the diagram.
 
@J.M. See also my answer here, with a little help from John Conway.
 
@BillDubuque Yes, I've upvoted it before. The "Gaussian trick" of pairing things up is remarkably handy.
 
@J.M. There I tried to emphasize the innate (group-theoretic) symmetry (pairing up reflections), but perhaps I didn't succeed.
 
@BillDubuque Hehehe pics are usually cool.
 
7:11 PM
@BillDubuque Why do you think so?
 
@RagibZaman Hey!
 
@PeterTamaroff Hey man
What's up?
 
@RagibZaman Just doing some topology, as usual.
And watching some sports.
 
Oh which sports?
 
Basketball and the Pre-olympic stuff
 
7:24 PM
Awesome! Do you normally watch any basketball?
 
@RagibZaman Only when Argentina plays.
 
Because subsequent answers to analogous questions often seem to miss the innate symmetry. But perhaps those answerers never saw that (or related) threads. It's important to teach students how to generalize basic problem solving methods, to realize that a "trick" may be the genesis of a beautiful theory. But, alas, the SE software encourages fast answers, to the detriment of teaching.
 
@RagibZaman I'm playing with balls my self now, =D
Actually with balls, closed sets and nbhds.
 
@PeterTamaroff I forget exactly what time but I think some time today Argentina plays the USA in an exhibition game.
 
And metrizable spaces.
 
7:25 PM
LOL oh ok.
 
@RagibZaman It is starting in a few.
I jsut watched the repetition against Spain.
 
@BillDubuque "a 'trick' may be the genesis of a beautiful theory." - that can definitely be a slogan for math... :)
 
@RagibZaman I'm also thinking about reading Prof. Scott's awesome answer here thoroughly and doing what he suggests.
It is such a great answer.
@RagibZaman I'm trying to prove that in a metrizable space, any nbhd of a point contains a closed set that is a nbhd of that point.
I wrote something
But it depends on the proof that $B(a;\epsilon)\cup \operatorname{Bdry}B(a;\epsilon)$ is closed.
 
@J.M. Quick question: if you take a beta blocker, how many days will it have a negative effect on your memory? I guess I could google it but if you happen to know that's a much faster way of finding out. : )
 
Which I don't have as a result yet
 
7:31 PM
@PeterTamaroff I've got to do something, I'll be back in 10 minutes. Perhaps Matt can help you if he wants before then, if not I'll help when I get back. See you all soon.
 
@RagibZaman The game will start in a few.
 
Ping me if you have an answer, otherwise never mind. : ) I shouldn't sit around in here, got to go study.
 
@MattN. Oh, just go and study. I'll try and manage.
I just look at some complements and intersections.
In fact, I think I got it.
Maybe.
 
@MattN. It's quite old. About a month or so ago.
 
7:48 PM
@PeterTamaroff Ok I'm back. If you think you've got a proof and want me to check it, shoot.
@PeterTamaroff I'm going to get back to sleep at 6 if you don't need any help right now.
 
@Gigili That is indeed old. Well, when times are bad they will always get better eventually, you just have to wait patiently.
 
8:03 PM
@RagibZaman What time you got?
Its 5 pm here.
 
@PeterTamaroff 6 a.m. here.
 
@RagibZaman AM!
Where are you!?
 
lol yea I live in Sydney, Australia
 
@RagibZaman Oh! We're soooooo far away, dude.
You haven't sleept yet?
 
@PeterTamaroff Yea, living in Australia, I'm used to being far away from almost everything.
 
8:05 PM
My back is hurting like I have seventeen backstabbing monkeys stab it with flaming knives!
 
@RagibZaman LOL
 
@MattN. Well, now that I think more carefully about it, it was next week or something that I changed the avatar! I hope so, too. Thank you for your attention!
 
@JonasTeuwen From what? Sitting with bad posture all day?
 
Would be cool if I could kill them all with one bullet.
 
@RagibZaman Not sure. I am Torbulator, the creator of eternal pain. No problem!
 
8:06 PM
lol
 
@RagibZaman I wrote this so far
"Let $N$ be a nbhd if $a\in X$. Then $O\subset N$, where $O$ is an open set and $a\in O$. Since $O$ is an open set of the topology given rise from a metric, it is the union of open balls. In particular $$O=B(a;\delta_a)\cup \left(\bigcup_{x\in O}B(x;\delta_x)\right)$$
Now choosing $\delta<\delta_a$ $B(a;\delta)\cup \partial B(a;\delta)$ is a closed nbhd of $a$.
But I don't have that theory at hand now.
 
Perhaps you can try flesh out this idea
 
@RagibZaman Uh?
 
If N is a nbhd of $a\in X$ then $O \subseteq N$ where $O$ is an open set containing $a.$ We can find an open ball around $a$ with a radius small enough so this open ball is a subset of $O.$
Set that radius as $r.$ Then an example of the desired closed set is the closure of the open ball with radius $r/2.$
 
@RagibZaman Right! But that's because $B(a;\delta)\cup \partial B(a;\delta)=\overline B(a;\delta)$!
Hehehehe
 
8:17 PM
I don't think the only way to show that this set is a suitable candidate to prove the theorem is to use boundaries (if that is the concept with the theory not at hand now?)
 
OK, I'll write it down-
@RagibZaman I just kidding
 
Basically this problem boils down to showing $ \overline{B(a,r/2)} \subseteq B(a,r).$
 
@Gigili But killing people usually won't help make things better. So I wouldn't bother.
 
@MattN. That's still an option when you can't do anything else.
As useless as deleting the problem, but still.
 
@RagibZaman Isn't that trivial?
 
8:22 PM
How to get the product of two polynomials? Through the opbvious way: (3 x^3 + 12 x^2 - 2 x + 1)*(5 x^3 + 3 x^2 - 18 x + 1)?
 
@PeterTamaroff Well it still needs a little bit of work to be formally justified, but it is intuitively obvious.
 
@GustavoBandeira Legend says there is something called the distributive laws of multiplication and addition. =D
@RagibZaman I did the following:
We have that $O$ is the union of open balls and in particular $B(a;\delta_a)\subseteq O$
 
@PeterTamaroff I know, but as I always forget them, I get em all in paretheses.
 
@GustavoBandeira OK. Start with $(a+b+c)(d+e+f)=a(d+e+f)+b(d+e+f)+c(d+e+f)$
 
@Gigili No, I can't empathise with this idea. Besides, if you kill someone you'll end up in prison. So instead of improving your situation you only make it worse.
 
8:26 PM
Choose $\delta<\delta_a$. Then $\overline B(a;\delta)$ is a closed neighborhood of $a$ for $B(a;\delta) \subset \overline B(a;\delta)$ and the closure of a set is closed.
 
@MattN. You can also suffocate them with their own pillow.
 
@Gigili And throw them in the Grand Canyon?
 
Not necessarily. Only if it's worth it!
 
@Gigili OK. Flesh eating bacterias.
 
@Gigili I don't understand you. You said you were a religious person. But afaik, religious books such as the bible or the koran explicitly say that killing is a sin. On top of that, aren't you supposed to have faith and trust in G*d to make things better for you?
 
8:29 PM
@MattN. Well, if they were good people they'll go to Heaven!
She's doing them a favour.
 
I see. I'm not sure this makes any sense.
 
@RagibZaman Then I have to prove that $\overline B(a;\delta)\subset N$.
Which of course follows by proving $\overline B(a;\delta)\subset B(a;\delta_a)$
 
@MattN. Pft. That's a meme. I wouldn't kill someone IRL. Everybody knows I'm not serious.
 
@Gigili Ah, good. Sorry for being slow. : )
 
@RagibZaman How would you prove that inclusion?
 
8:34 PM
@PeterTamaroff I'm going to work with the sufficient case of showing $\overline{B(a,r/2)} \subseteq B(a,r)$ ok?
 
Hi guys. Quick q -- when using IM etc, is there a good online pastebin that will render mathjax?
 
If you still want to use general deltas I'm sure you can adapt the idea.
 
@RagibZaman Well my $\delta_a$ would be your $r$ and my $\delta$ would be your $r/2$.
 
Just intuitively, if $x\in \overline{ B(a,r/2) }$ then $d(a,x) > r/2$ is impossible.
 
@RagibZaman Thus $d(a,x)<r$
 
8:37 PM
yup
 
But we also have to take care of the boundary points.
Those are precisely the points such that $d(a,x)=r/2$ methinks.
 
The points such that $d(a,x) = r/2$ AND are in the closure, are the points on the boundary, that is all we can say from thinks we have proven.
 
@RagibZaman There is some word missing there.
 
Ok when we showed $d(a,x) > r/2$ is impossible, we're already done, no?
 
@syrion Did you Google it?
 
8:40 PM
@RagibZaman Maybe?
Maybe I can use that
I have the theorem that $F$ is closed in a metric space $X$ iff for each $x\in X$; $d(x,F)=0$ implies $x\in F$.
 
We don't need to do such things, this is simpler than we are making it out to be.
After we show $d(a,x)> r/2$ is impossible, we deduce $d(a,x)\leq r/2 < r$ so $x\in B(a,r)$ and we get the subset statement.
 
@Gigili: I did before asking, saw nothing, and then immediately after I realized what I should google for and found it :-\
 
@RagibZaman Yikes. I need to think about stuff in a simpler way.
 
@syrion Haha. Googling "online TeX renderer" yielded some results.
 
Ali
I'm gonna use the following statement in a research paper and am just wondering if it sounds mathematically correct.
"As $u \to \infty$, this chance monotonically increases while being upper-bounded by a horizontal asymptote."
 
8:45 PM
Yeah, I ended up at mathb.in
 
@syrion FWIW rob's sript works both in Facebook and Google Talk
 
@PeterTamaroff Indeed, for example you seem to use the idea that open sets are unions of open balls almost every time you can, I implore you to place that in the back of your mind, it is far less useful that it seems.
 
@Peter I don't know what that script is.
 
@RagibZaman OK. You go to uni with Benjamin right?
 
This is my first time at math.-, I'm starting to try to learn category theory so that I can do functional programming more effectively
 
8:46 PM
@PeterTamaroff No, he goes to Australian National University in Canberra, I go to the University of Sydney, so most of the year we are in different cities.
 
I'm borderline innumerate so it's proving bumpy :-\
 
@syrion Good one. The one I found doesn't look that good.itools.subhashbose.com/educational-tools/…
 
@RagibZaman Oh OK.
 
@PeterTamaroff So let us just see how simple this problem can be:

We want to show that inside every nbhd, there is a closed nbhd of a point $a\in X.$ We draw a picture. We know we can draw an open ball around $a$ and a good candidate to be a closed nbhd would be the closed ball with half the radius, but we need to show that is indeed a subset. So we need to show $d(a,x)> r/2$ is impossible for points in the closure of $B(a,r/2).$ *Back to this in a second. With that, $d(a,x) \leq r/2 < r$ so $x\in B(a,r)$ as required.
@PeterTamaroff Anyway, it's almost 7am here now, I want to get at least 3 hours sleep before I start my next day lol. Good luck with your studies and even more so, your countries game against the US. Gnite all.
 
@RagibZaman Dude! Sleep!
 
9:17 PM
so what is the difficulty today @PeterTamaroff?
 
@DavidWheeler None.
For the moment. ^_^
@DavidWheeler How are you?
 
@J.M.: I have a very small, but neat set of functions for computing the Euler-Maclaurin expansion that I used in this answer.
 
i am fine
 
@robjohn Hmm :-). What was your first motivation for studying this? It is not really very harmonic analysisy or perhaps I don't really know about all of it :-).
 
@JonasTeuwen I have studied a lot of things since grad school. Asymptotic expansions are pretty cool.
 
9:32 PM
@robjohn Yep! They really are. You have driven my to studying them 8-). What is your cutest reference for them?
@robjohn But did you just study them because they are cool or did you want to solve a special kind of problem?
 
@robjohn that looks like the kind of thing where you could spend a week, and eventually say: i did it! the answer is 4!
 
@JonasTeuwen I actually started studying them when I had a program to test out a random number generator that set a random pixel on the screen. I wanted to find out how long it should take to fill a screen of $n$ pixels, and I computed $n\sum\limits_{k=1}^n\frac1k$ iterations.
 
@robjohn Ah :-).
 
@JonasTeuwen Then I started looking at the asymptotic expansion for $H_n$ and found the EMS, etc
 
@robjohn For Hermite or the harmonic numbers? I love Hermite 8-).
 
9:36 PM
@JonasTeuwen Harmonic numbers ($n\sum\limits_{k=1}^n\frac1k$)
That's a reference to the message above, I know that $H_n=\sum\limits_{k=1}^n\frac1k$ :-)
 
@robjohn Hermite. They rule.
@robjohn Have you heard of "binomial type" functions/operators?
 
@JonasTeuwen What are they?
 
I had defined them and made some theorems, but some nasty combinatorist did that already in the 60s and 70s :-((( 8-).
@robjohn A special kind of polynomials.
 
@JonasTeuwen The definition there uses $p_n$ to define $p_n$.
 
@robjohn Mmm dangerous 8-). Let me see.
@robjohn I think it does not. You have a sequence $p_n$ say the "probabilists's Hermite polynomials" and then you can verify they satisfy that condition. What do I miss?
 
9:48 PM
@JonasTeuwen They look similar to these
@JonasTeuwen Ah... they satisfy that equation. It looked as if that was supposed to be a recursive definition. Sorry.
 
@robjohn Mm, they are certainly not unique :-).
Perhaps we can actually generate them that way... must think about that. Good question, thanks for that!
@robjohn Combinatorical polynomials? Did you think of that name or is it something known?
Do you know what kind of combinatorics this Bourgain dude uses wrt harmonic analysis?
 
@JonasTeuwen Combinatorial (no "c") is what I call them since they are simply $f_k(x)=\binom{x}{k}$
 
@robjohn Yes, Pochammers. (and thanks)
 
@JonasTeuwen nope
@JonasTeuwen divided by $k!$
 
@robjohn Oh :-). Is this your theorem?
 
9:52 PM
@JonasTeuwen I am sure it was not original, but it was new to me.
 
@robjohn Good.
I have that all the time.
Without knowing the stuff a priori, so maybe I am just born too late 8-).
@robjohn Why do you plain text it?
 
@JonasTeuwen I was answering a lot of questions about showing that some polynomial was always divisible by some integer...
@JonasTeuwen This was done back before your time, when LaTeX was not so easy to do on the net, and all we had were monospaced fonts. :-)
 
8-).
 
@JonasTeuwen: have you ever seen any way to compute $\sum\limits_{k=1}^n\tan^2\left(\frac{k\pi}{2n+1}\right)$?
It came up in an answer to a question, and there were a couple of ways presented.
 
@robjohn Hmm... No. Some Poisson summation probably no good eh... too finite.
@robjohn You mean you found a couple of ways to compute it?
 
9:58 PM
@JonasTeuwen Yeah, the finiteness stops Poisson
 
Martingale stopping theorem to the rescue!
 
@JonasTeuwen Actually, yes. I had found one, only to see that Joriki had also done the same answer before I posted, so I came up with a second way.
 
I wouldn't know from the top of my head. Needs some incubating.
@robjohn Do you haver a link please? :-).
 

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