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21:00
@TedShifrin: I'd appreciate that. My ideas are still revolving around the topology example I gave you.
Nope, Mike. But I have a good guess that it's completely dual to the $\Bbb CP^n$ story.
@Balarka: Here's my hint: Radius of convergence.
I don't know that story. In that case, you're constant on complex planes but then quarter-pinched positive in general?
@MikeM: So I'm guessing that with holomorphic sectional curvature $-1$, the real sectional curvatures vary between $-1$ and $-4$.
I guess there's some linear algebra (polarization stuff) that explains the factor of 4?
$1\le K(\Lambda)\le 4$ in the projective case.
21:02
maybe it realizes that on the totally real planes?
Well, we can get it all out of Lie algebra stuff if we want, @MikeM.
totally real is the wrong word...
@TedShifrin Oh, pointwise, because of global symmetry... sure
i then also wonder what the story is for $\Bbb{HH}^n$, which is godawful notation
For a symmetric space, curvature comes immediately out of the adjoint representation of the right thing on the right thing.
technical terminology, that :P
Yeah, that story I know nothing about. I've done the projective space computation a few times in my life.
21:04
beautiful classic theory i know nothing about.
i skipped those chapters of besse (:
I can send you homework problems, Mike :P
no time...
I know.
You have no time, and my brains are disappearing :P
3
spent 2 weeks fixing a lemma that I thought was an easy fix
got it a couple hours ago though
Great :) Mazltov.
I thought my whole thesis had fallen apart after I already had gotten a job. You will have occasional panics in your math life.
21:06
shana tova
maybe i'm a bit early
L'shanah tova!
My more observant FB friends are doing the rosh hashanah thing. But I'm an atheist non-observer.
Last night, I saw one of my former students putting together a party in a classroom for the Jewish students union.
@Semiclassic @Balarka et al : Fantastic question.
let's say I'm an atheist observer
I'm an atheist, but I'll still observe that people are doing it :)
21:08
I'm not Jewish but it looked like it was going to be fun - I would have liked to have been there
I'm sure you could have asked to crash the party, @ALannister.
Parties are fun.
probably ;) That girl was awesome and she just loved me.
this year so far I'm having trouble learning people's names.
i always try to do that in my recitations because I think it's important. Some other TAs don't care.
21:10
I'm finding it hard to understand the conditions on velocity, though
oh, bah, I was misreading it
It is important, @ALannister. I always tried to know all my students' names within the first week or two.
I hadn't decided yet if they had the sign right, @Semiclassic.
are we treating the trajectory of the plane as a plane curve? otherwise I don't know what it means to enclose area
heh. the plane's trajectory is a plane curve :P
Yes, it's a plane curve. But I think they have the sign wrong on the $\pm a$.
Also, the constraint should be pointwise on the air speed, not on its integral.
21:12
right.
I guess that, since the plane's speed is constant, we can assume it to be whatever we want
so we can take the curve to be arc-length parametrized
Oh, I guess they're right. The actual (ground) path is $(x(t),y(t))$, so the ground velocity is $(x',y')$ and the air velocity is $(x',y'-a)$.
No, you're wrong on the arclength parametrization. The ground path won't have constant speed.
@TedShifrin I'm too tired to think today
i'll probably go to sleep
21:15
G'night, @Balarka.
would that work in the reference frame of the air?
@TobiasKildetoft more hints please...
Don't give me non-inertial frames :)
@Semiclassical It is interesting that by rearranging the letters of hmm, we get mhm.
uh
air speed is constant
21:16
All right. I gotta go do stuff.
nothing noninertial about htat
But the plane still accelerates, @Semiclassic. Duh.
@BalarkaSen See you in your dreams.
@TedShifrin could you point me in the right direction: let p divide |G| but not p^2 divide |G|; let H=p-subgroup; prove N(H)=C(H)
@Leaky: I'm thinking about something else now, evidently.
21:16
I'm confused now.
Constant speed ≠ constant velocity!!
@TedShifrin sorry
Hi @Ted
ya. and? the condition for arc-length parametrization is that ds/dt = 1.
Hi @Astyx.
Semiclassic. I think this is totally not relevant, anyhow.
21:18
so if we're in a reference frame where the plane has constant speed then I'm not seeing the issue
I'm complaining that it's still an accelerating frame, but maybe it's no matter.
I'm not saying to take the plane's reference frame. I'm saying to pick a reference frame that moves with the fixed air velocity $a$. That velocity isn't changing and so there's no acceleration.
Actually, there's another boundary condition they left out. The curve has to close up in time $T$.
Ohhhhh, I did not understand that you meant that.
I think we want the usual ground frame so that we can actually measure area of the curve.
21:19
yeah, that could be the fly in the ointment
Which is the fly in which ointment?
the area of the curve being best described in the ground frame
Right.
whereas the plane's trajectory is most easily described in the air frame, where it's got constant speed
thing is, if it's a closed trajectory in one frame it's closed in the other as well
and I want to say (perhaps wrongly) that the areas in the two frames are related quite simply
wait.
hrm. no, I'm wrong
if I throw a hammer through the air, it'll rotate around its axis in its frame of reference
but relative to the ground, it's most definitely not making a closed loop
okay, there goes that thinking. oh well
back later
LOL, bubye.
21:24
That's what the doctor always says to me, bubye, lol.
Mathematically, it's something like $\int_C x\,d\tilde y = \int_C x\,dy \pm \int_C ax\,dt$, but $\int_C x\,dt$ certainly needn't be $0$. @Semiclassic
@Leaky: By $N(H)$ you mean the normalizer of $H$. What's $C(H)$? I know about centralizers of elements.
@TedShifrin centralizer of subgroup: $C(H) := \{g \in G \mid \forall h \in H: gh=hg\}$, i.e. elements that commute with every element in subgroup
Oh, the intersection of all the centralizer subgroups.
You obviously need an appropriate group action. Act by $G$ on the set of subgroups conjugate to $H$ or something?
I last taught algebra 5 years ago, so I don't know this stuff immediately. I'll stick to diff geo :P
Did your stuff in algebraic geometry not involve a hefty amount of algebra and the like?
This isn't quite inspired by your previous comment but I was wondering this for a while, I know there's the whole super abstract stuff but how much of a difference is there in the more standard stuff?
@Ted
21:43
i suspect Ted's algebraic geometry used less of what people might call hefty algebra than my current project does, and i'm only using it while glaring at it
What's that about?
@Semiclassical did you manage to finish the integral I told you about? (I think I encouraged you a lot, and that should help to some extent)
(I think @Semiclassical just left the chat to work on the integral in complete silence)
Okay, let me retype it.
$$\sum_{k=1}^n k^{2p+1} \equiv \left(\sum_{k=1}^n k^p\right)^2 \ \text{(mod p)}$$
22:00
@Waiting I did not think he likes integrals.
This evening I received this question from a student. It's very nice to try, and the proof is short.
@Jasper @Semiclassical is a great fan of integrals actually (as far as I know).
@Waiting I was only interested in integrals in high school.
@Jasper And then what happened?
@Waiting Then I got more interested in other kinds of math, lol.
@Jasper Like? Algebraic geometry?
22:05
@Waiting Well, like maybe just analysis itself without computing hard integrals. =D
@Jasper :D
@Jasper Computing (hard) integrals is actually an art, an addictive one.
@Waiting Yes, it is. It's something I can't do, just like I can't draw.
@Jasper Does it seem a too difficult task to do? Or how do you perceive it?
@Waiting Well, I am not as talented as you, so it is very hard for me.
@Jasper This makes me sad to some extent. I told you many times it's not about being talented, it's about very hard work and mindset, the attitude you have in mathematics.
22:12
@Waiting Hmm, I will try to see if that is true. =D
@Jasper That sounds good!:-)
@Waiting May I ask again about the status of your book?
@Jasper In progress.
@Waiting Are your math degree plans also in progress?
@Jasper There are always plans, but not to share here. ;) I feel this place (chat) is still dark, cloudy and hopeless but maybe is just a feeling.
22:14
@Waiting Hmm OK. If you want to share, you can email. =D
hello
new here
i wanted to ask are these rooms private or something?
@Jasper OK
@Pookaros Every room can be read by everyone in the world.
@Jasper ok ty
@Jasper does it bother u joining ur integral-talent discussion?
@Jasper for instance, this question could be finished in two lines
64
A: How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

CleoIndeed, there is a closed form for this integral: $$I=\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).$$

22:18
@Pookaros You can talk to everyone here. =D Of course it does not bother me.
@Jasper it is precisely so.
Let me share one more
@Waiting can this be solved in complex space using Res?
15
A: Closed form solution to $\int_0^1\arctan^2(x)\,\sqrt{x}\,dx$

Cleo$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$

@Pookaros It could be, but I did it using real methods.
@Waiting Well can this result from $e^{iz} + e^{-iz}=2sinh(z)$
@Pookaros I would better use here a nice series representation of $\arctan^2(x)$ in terms of harmonic numbers.
(which is easy to get)
22:25
@Waiting hmmmm yeah i see ur point.
It's nice when you look over the closed-form and figure out what happened during the calculations.
I have to leave now.
Nice integral to try, just for fun.
@Waiting i will mate bb :)
it turns out that proving "n^2 is even => n is even" takes much more work than I would expect
it's amazing how many things I take for granted
@LeakyNun Is that a recent photo or an old one?
@Jasper half a year ago
22:35
@LeakyNun Hmm, you look quite short to me, am I right?
yes you are
I am quite short as well, 1.66 metres.
0
A: Prove if $n^2$ is even, then $n$ is even

Kenny LauDefinitions $x\text{ is even} := \exists y[y+y=x]$ $x\text{ is odd} := \exists y[y+y+1=x]$ Axioms $\forall x[S(x) \ne 0]$ $\forall x \forall y[S(x)=S(y) \implies x=y]$ $\forall x[x+0=x]$ $\forall x \forall y[x+S(y)=S(x+y)]$ $[\varphi(0) \land [\forall x[\varphi(x) \implies \varphi(S(x))]]] ...

It's 6:41 AM now
if I were to complete my proof
I wouldn't be able to sleep before 9AM
so let's just leave it there
23:16
Guys, what does it mean for the quotient group to set certain values equal to 0 modulo the equivalence relation?
23:46
it means those 'certain values' are in the same equivalence class as 0 under the relation
So is a quick way of forming the quotient group $G / N$ just to set the generators, and thus their multiples, of $N$ to 0?
@LeakyNun what an overkill
Anyone? :P
if you interpret "multiples" appropriately, I suppose, @gian
ordinarily it's the normal subgroup they generate

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