speaking of which, i need to hit the bed.

@Balarka: Ahem!

@MikeM: so what is her question about, then? :)

@anon While you're at it, try to prove that if $X \stackrel{p}{\to} Y \stackrel{q}{\to} Z$ is a bunch of galois covers such that $r = q \circ p$, then there is a short exact sequence of deck transformation groups $1 \to Aut(p) \to Aut(r) \to Aut(q) \to 1$. i have probably posted it here more than once, but it's something really nontrivial I actually found when studying covering space.

hi @anon ... Thanks to you, I got another badge :)

you're welcome. that will be $2.

also, note the similarity with the short exact sequence of galois groups.

People tend to overreact around here ...

@TedShifrin wanna fight about it?

Yes, that would be fun, @anon :)

Pitchforks anyone ?

@BalarkaSen interesting

it is.

@anon A pastor told me once to try this: Jeremiah 29:13 "You will seek me and find me when you seek me with all your heart.". Don't imagine I'm powerless in finding flaws in the Bible and I'm incredibly hard to be made to believe in anything. I was maybe less then 10 years olds when I was contardicting people on religious themes.

Sometimes claims are false.

note that setting $X$ to be the universal cover, $Y = X$ and $Z = X/G$ gives back the "homotopy short exact sequence" $$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$$

than

@Ted: I answered the question. The McDuff question is either false or trivial.

That doesn't seem to be the usual homotopy sequence of a fibration, @Balarka.

any easy-to-visualize examples of not-regular coverings?

it's a special case, @Ted

$G \hookrightarrow X \to X/G$ is a fibration

$G$ is a discrete group here

Ohhhh, if $G$ is discrete, then it is fine.

oh, sure $G$ acts prop. disc. and freely on $X$.

that's a covering map, so saying fibration is overkill....

i was referring to Ted's "homotopy sequence of fibration" thingy, just to demonstrate that it follows from the long exact sequence

yeah, I was being a dope, @MikeM

Good thing Ted retired.

@anon But I prefer to check, to test, not to take for granted what other told me only because, say, Stephen Hawking is an atheist or many other famous scientists that have embraced the atheism. Besides that, life experiences can have a huge impact on your beliefs. Nowadays, the easy way is to be an atheist and you're done, no headache at all with this kind of questions.

@MikeM: BTW, for us non-symplectic geometers, you should define Liouville vector field in your post.

Being an atheist is much easier nowadays, but it's only "the easy way" in certain liberal social settings. Go to the Bible belt or (far worse) any theocracy and the story is much different. Atheism is much more available now, and there are vocal proponents, so that shallow acceptance of perceived authority is a phenomenon no longer as restricted to religion.

Just, for example, so that one understands that the condition is not invariant under multiplication by $-1$.

@Ted: I'll do that when I get back to campus. You can feel free to do so if you want.

@anon can't whip up a good geometric example right now, i guess i am too sleepy. but you can always begin with some space with nonabelian fundamental group and construct up a covering space corresponding to a non-normal subgroup

sorry, gotta go sleep.

@MikeM: You say the. Is it clearly unique? That would surprise me. What if there are lots of symplectomorphisms?

the word "symplectic" always sounds like stirring a bowl full of soup to me.

GOOD NIGHT @Balarka.

lol.

@Ted: That's what upsets me. It is not clearly unique. It is clearly not unique. But this is the phrasing that's used everywhere.

I was about to edit, and didn't want to write garbage. I would say "a" vector field.

@anon Kind of true.

Well, there's a unique one up to scalar multiples. But why can't we add the generating vector field of a symplectomorphism, if there is one?

I would edit that myself. The point is that standard phrasing is confusing.

A unique what up to scalar multiple? That's false. There are plenty and scalar multiplication does not preserve being Liouville.

Right, that's my point. If $X$ is Liouville, $cX$ is also if and only if $c=1$.

Ok. I will edit it later. Again, not at computer.

But I can add any $Y$ which generates a symplectomorphism.

@Chris'ssis there are many countries like sweden though, or say to china, in which atheism will also be pretty easy (although I don't think that's a recent thing in those kinds of areas).

Give me 1 hr, then unload on me. :D

Nah, I'm not criticizing.

You're probably texting in class ... :D

@Chris'ssis Any elegant way to get a formula for $\sum_{n=0}^\infty\dfrac{1}{(a_1n+b_1)\dots(a_kn+b_k)}$ ?

@anon Or Czech Republic

(at fixed $k$)

Can anyone say what's on the paper in this pic? : scontent-ams2-1.xx.fbcdn.net/hphotos-xta1/v/t1.0-9/p320x320/…

@jarlemag a pen :P

Ahem, that's beside the paper...

ink?

No, @jarlemag. Some inverse Fourier transform.

Japp chocolate bars ?

We're getting closer!

Yes, I thought so. But wasn't entirely sure there wasn't something else using similar symbols.

What about the next lines, any ideas?

Can't read 'em.

@anon If I wanted to be an atheist it would be so easy, I read the first few pages from the book of genesis and I was done. Some of the experiences in life can change the perception on God, that's definitely true. I don't know if reading the bible alone could have ever made me seriously think of the existence of God.

Bigger version here: scontent-ams2-1.xx.fbcdn.net/hphotos-xta1/v/t1.0-9/…

Total nonsense.

Hehe, OK.

That people interpret their life experiences through the local or familiar religious lenses strikes me as more the power of suggestion and salience than anything else. For instance, I am sure many Muslims believe their life experiences move them towards Islam and are equally as justified in that belief as Christians believing their life experiences move them towards Christianity. I cannot tolerate the arbitrariness, myself.

It has the appearance of a Laplace transform, but I don't know much about those, sadly

@anon I was expecting this kind of reply though. I understand you perfectly, if I were you I'd probably say the same thing.

pjs36: Do you mean the top line?

Yeah, the calligraphic font with an inverse, and using curly braces around the input which is a function... but again, that's definitely not my area of expertise

The script $F$ is ordinarily a discrete Fourier transform. A script $L$ would be Laplace.

I thought as much, but I really haven't seen a Fourier transform in years, and debated whether they had the same notation

Ted Shifrin: That was my understanding too. Studied both Fourier and Laplace transforms in Uni, but that's a while ago. :)

Still, I wasn't sure if the notes actually make mathematical sense...

Not that it matters much, was just curious how much effort they actually put into the ad.

@TedShifrin buon pomeriggio amico!

Wow, I am violet/blue today

Quite purple, yes, @Stan.

So, are you familiar with the KKT conditions?

not offhand

Is that just an econ thing?

@Hippalectryon hmmm, I don't know at the moment :|

what's that short for, @Stan?

@Chris'ssis I was doing $\sum{\dfrac{1}{(8n+1)(8n+5)}}$ so it got me wondering about that

Ok, @Ted, what do I need to fix? Add a definition of Liouville field, change "the"?

Yes, @MikeM.

@Hippalectryon presumably partial fractions and then relate it to digamma. one should then ask about $\sum_{n=1}^\infty\left[\frac{\alpha_1}{n+\beta_1}+\cdots+\frac{\alpha_k}{n+ \beta_k}\right]$ with $\alpha_1+\cdots+\alpha_k=0$

Did you get through the rest?

Not carefully ...

@TedShifrin Karush–Kuhn–Tucker conditions

@anon Integrals

I figured it was part of Kuhn-Tucker, @Stan. Dunno.

@Hippalectryon This is related to BBP-type formula, isn't it?

Hello @jarlemag

@Chris'ssis BBP formulas also have an exponential factor in the summands

@TedShifrin Alright, just checking/curious.

although I guess base=1 could be considered a special case

@Chris'ssis Maybe, but it's not meant to be.

@TedShifrin So when you teach / taught students, do/did you encourage students to collaborate?

And what do you think counts as "collaborating"?

Absolutely, @Stan, but I ask that they write things up individually.

Yes, I never collaborated in high school, but I am doing more of it now and I feel like I am learning much much more.

Actually, SE helped me see how useful it is, so I've gotten better at it.

@anon Yeah, I know, but it made me think of them for some reason, I think I saw a series representation of $\pi$ that contained those fractions.

@Hippalectryon How you did that series (btw)? You have some tools to tackle it, of course, no concern therefore.

@Chris'ssis Integrals

@Hippalectryon OK :-)

@Hippalectryon You can look at it using digamma function also.

@Hippalectryon I think I can do that genneralization of yours using digamma function.

If you find a nice form, do tell me ! :-)

@Hippalectryon If I do that, then I write a paper first! :D

Sure :D

@Hippalectryon You're so funny sometimes! :-)

Uh... thanks... I guess ;-)

@r9m Sure. In any case, everything here is usable with proper references.

@robjohn :)

@robjohn I think $ \displaystyle 2^{-n}\binom{n}{k}\sim\frac1{\sqrt{\pi n/2}}e^{-2(k-n/2)^2/n}$ is efficient for about $n^{2/3}$ central terms (i.e., in the range $n/2 - \mathcal{O}(n^{2/3}) \le k \le n/2 + \mathcal{O}(n^{2/3})$), however that does not affect the final result :) (the remainder goes to 0 once we divide by $n^p$ and take the limit)

@Chris'ssis It's funny how wolfram can give two really different forms for the same answer

@Hippalectryon you mean they are different numerically speaking?

@Chris'ssis No, just the form in which they are computed

@Hippalectryon Ah, yeah, indeed. :-)

@Ted: Updated.

@r9m I'll have to estimate the sum of the binomials with the cumulative normal function, to get an estimate of the error.

I made significant edits that should improve readability.

@robjohn I just had a heuristic idea .. I don't know how to get the error terms for such instances

I saw, @MikeM ... Heading out in a few to take a friend to dinner for his belated birthday.

I'll have to digest your answer to your own question later tonight :)

@Chris'ssis Digamma definitely seems like too big guns here though (for the simple one wih two terms on the denominator)

@Hippalectryon Sure. I was referring to the generalization.

@Stan: Particularly with hard classes, having students work on stuff a while individually and then get together (in small groups) to brainstorm ultimately results in a lot more learning and a lot more accomplished. :)

Yes, so far I have found that the optimal group size for problem sets is 3-5 and the optimal group size for studying for tests and reviewing material is 2.

@TedShifrin I tried discussing the material with 4 people (including me) and it just became to opinionated and people couldn't agree quickly enough to learn.

LOL ... I expect a complete economist's analysis, with all statistics analyzed :P

It's late ... (I don't know how the last 3 hours went so incredibly fast)

(hmmm, I was planning to finish some more proofs)

@TedShifrin LOL yes, and then I should propose solution using the method of Lagrangian multipliers for the optimum utility maximization given a learning quantity constraint.

No, you're trying to maximize learning :P

Hahaha yes, exactly.

although, maybe learning isn't my highest priority :p

Of course not. Grades are ...

Mmmm....no I care more about learning than grades. Grades are a necessity for jobs, but learning is a necessity for life and enjoyment.

At least for me.

Oh, so impressing the right friend is the highest priority :D

Grades aren't that much a necessity for jobs. To some extent for grad school, but even that takes a grain of salt.

Grades are a necessity to get in a better school :/

@TedShifrin So how much space does 400 books take up? How many books is that in units of bookshelves worth of books?

For grad school, only grades in one's subject really matter ... and, ultimately, in the US, GRE scores and letters of recommendation are most important, as long as there are good grades.

@TedShifrin That's in the US

Doesn't apply for competitive exams :/

@TedShifrin When you say GRE, do you mean math subject GRE?

LOL ... yes, @pjs36

OK, I'm leaving to go to dinner. Be bad, y'all.

I just wanted to be clear! I figured it was an LOLable question :P

@TedShifrin Will do, as always :p later

@anon only one thing before leaving (since I didn't manage to do much math now) - when you'll see someone with ugly squint healed after the pray of a Christian, I mean someone you know very well, from your neibourhood that you knew for many years, then you'll realize that life experiences can change you. This is only a tiny example.

@Hippalectryon That can be seen to be $$\frac1{16}\left[e^{3\pi i/4} \log(1-e^{\pi i/4}) +e^{\pi i/4} \log(1-e^{3\pi i/4}) +e^{-\pi i/4} \log(1-e^{5\pi i/4}) +e^{-3\pi i/4} \log(1-e^{7\pi i/4})\right]$$

I'm out.

@robjohn What way led you to that form ?

@Chris'ssis good night

@robjohn Good night.

(and to all)

@Hippalectryon Writing the repeating terms $(0,1,0,0,0,-1,0,0)$ as a sum of eighth roots of unity

What repeating terms exactly ? (we're talking about $\sum _{n=0}^{\infty } \frac{1}{(8 n+1) (8 n+5)}$ right ?)

@Hippalectryon $$\frac1{(8n+1)(8n+5)}=\frac14\left(\frac1{8n+1}-\frac1{8n+5}\right)$$

Oh, those terms

Yeah, that's a nice way :D