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17:05
@anon: I set out to prove you wrong and discovered that $S^2 \times S^2$ double covers $\text{Gr}(2,4)$ which is pretty cool
(pretty sure you're right)
other than that I don't have anything intelligent to say.
hmm, what's the cover?
pull back the $S^2$-bundle along the map $S^2 \to \Bbb{RP}^2$ (assuming it exists, which I believe)
so you get an $S^2$ bundle over $S^2$, of which there are two; one is $S^2 \times S^2$ and the other is $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$. i can rule out the second one with some fiddling that I don't want to write down.
i would bet you can come up with a more geometric description of this
hmm
for both this idea and your question maybe try thinking of Gr(2,4) as a homogeneous space O(4)/(O(2) x O(2))
Yes, @anon @MikeM, one of my favorite math facts it that the oriented Grassmannian $\tilde G(2,4)$ is diffeo to $S^2\times S^2$. You can do this by linear algebra with the Plücker embedding in $S^5$ or by looking at the Hodge star operator.
Since $\tilde G(2,4)$ double-covers $G(2,4)$, your stuff should follow from this.
17:18
supposing $\widetilde{\rm Gr}(2,4)\approx \Bbb S^2\times \Bbb S^2$, how does this give the bundle $\Bbb S^2\to{\rm Gr}(2,4)\to\Bbb P^2$?
I've actually never seen your "fact" before, which makes me slightly suspicious, but presumably we have to chase the $\Bbb Z/2$ action through everything. I'll try to write something down in a bit.
you have a map $\tilde{Gr}(2,4) \to S^2$. you just need to show that it sends oppositely oriented 2-planes to antipodal points on the sphere to descend downwards
aka what ted said
One can write down the diffeomorphism quite, quite explicitly.
I'll get back to you anon, @anon. :)
I've been wanting to say that for a year. :D
okey dokey
i had also labored under the belief that if the total space is oriented and the fiber is connected and oriented, so is the base. this is obviously false in retrospect but jeez
17:22
No, all sorts of homogeneous spaces are counterexamples.
Hmm, maybe.
i mean, the hopf map gives you a circle bundle $\Bbb{RP}^3 \to \Bbb{RP}^2$. that's the first and most obvious counterexample
I have one doubt regarding one of the unanswered questions here on stack , since posting the question again would mean duplicating , what should i do ?
Hi
Suppose I have two shapes: A circle and a square described by

$$x^2+y^2=r^2$$

and

$$\{(x,y): |x-a|\leq s, |y-b|\leq s\}$$

What function can allow me to count the number of intersections for some given r, a,b,s?
...which of course is O(3)/SO(2) like you suggested :p
@anon, @MikeM: As I suspected, when we switch orientation on the $2$-plane we get the antipodal map on both spheres.
So this makes me suspicious about getting an $S^2$ bundle.
17:36
In any case, my initial curiosity was about which planes are stable under a given 1-parameter group of isoclinic rotations, so essentially the fibers of Gr(2,4)->P^2.
So what is $S^2\times S^2$ modded out by $(-1,-1)$?
@Ted: All we needed was a map $\text{Gr}(2,4) \to \Bbb{RP}^2$. Local triviality of this map follows from the local triviality above (or Ehresmann's if you like). Then you identify what the fibers are. They're surfaces with trivial fundamental group...
I said a few things. Which why?
The last one. I know the rest. And I should know exactly what this bundle is, but I haven't thought about it before.
It feels like the fibers are $\Bbb P^2$. I guess I need to write things down carefully.
17:44
@TedShifrin: You might be right, I made a calculation error. But the fibers would be two copies of P^2, not one.
hi @ted I didn't do your vectors yet I was on holidday
LOL, hi, @Ramanewb. You're always on holiday!
@ted Not that much...
@ted Can you help with this ?

I know $AB = AC = 14$
$M$ can move along $(AB)$
$N\in (CB)$ and $P\in(AC)$
$AMNP$ is a rectangle
Let be $AM = x$

How can I know in which interval x varies ?

pasteboard.co/1OyL2t4N.png
@MikeM, are you sure? I don't see that from the homotopy sequence, or intuitively.
hi @TedShifrin
17:48
Hi @Balarka
i haven't done much math today. sorry. :( i am a bit sick.
@Ramanewb, where are you stuck?
@Balarka: You deserve to take a break. No need to apologize. But I wish you'd stop being sick all the time!
@TedShifrin: Nope, oops again. I shouldn't be doing this anyway, but repeated mistakes makes a good excuse to stop.
@BalarkaSen Thank you for the help, I am humbled by your patience.
@BalarkaSen No break from doing math, that's the right philosophy.
17:49
@ted I don't really know what to do, and don't even understand how x could be an interval and not an unique value
LOL, @MikeM. It's actually a good question. I'll think some more.
I agree... but I have my homework.
I'm not criticizing, @MikeM.
@Ramanewb: What picture have you drawn? What do you know about $\triangle ABC$?
I wanted to post an easier version, but wait ...
@ted I forgot to say $ABC$ is rectangle in $A$
17:51
You don't need to say it. It follows. You mean (in English) $\angle A$ is a right angle.
@TedShifrin Well, it's a day wasted. I get sick every time the weather changes. The winter's coming, apparently.
@OverlyExcessive Anytime.
@Balarka: Have some soup and just rest and watch a movie.
good idea, i guess. thanks.
maybe i can watch your calc lectures instead of movies though :D
That is not what I meant.
Why do that when Harold and Kumar go to White Castle is so easily available?
17:57
Oh? Never even heard of it. I was thinking of watching this
@MikeM @anon: Since Mike's still here, I'm quite convinced it's a $\Bbb P^2$ bundle over $\Bbb P^2$.
I'll write out more later.
@BalarkaSen: Maybe wait a few years.
aight
I just watched Blood Simple last night. That'll be good for Balarka :P
@MikeMiller hm, ok.
17:58
I can't be recommending immature movies to people so young.
@MikeM, @anon: A good exercise is to check out the homology of $G(2,4)$ (which is known) with the appropriate Leray spectral sequence for the bundle :P
One needs to understand the $\Bbb Z/2$ action, of course.
<--- confused about spectral sequences
@TedShifrin: Remember you need a condition on the... darn
Go to bed, @Balarka. Now.
But I'm having dinner!
18:00
@TedShifrin: One should probably take $\Bbb Z/2$ coefficients to get away with our sins.
Actually, I guess if the fiber is P^2, it doesn't matter at all.
Excuse me genius people , can i ask for a small favour ?
Yes, absolutely, @MikeM. And it's only with $\Bbb Z/2$ coefficients that I know the answers, I think :) I usually live in complex Grassmannians :P
Very few genius people here, @Sujith. What's up?
@TedShifrin: It can't possibly change the homology of the fiber is P^2, so nothing matters...
But, @MikeM, we should be able to decide easily whether it's $S^2$ or $\Bbb P^2$ that's the fiber!!
Could you please give me an hint on how to approach this math.stackexchange.com/q/1489109/262120
Im not able to picture the isoceles triangle with a rectangle ...
18:04
I don't even understand the wording, @Sujith. Do you/they mean the square of the distance is the area of the rectangle?
@TedShifrin You expect to understand wording of a geometry problem from an Indian textbook? Pretty much hopeless.
Yeah, @Balarka. Very upsetting.
Such is our country.
Besides, @Sujith, there is no rectangle. The line segments from $P$ to the legs of the isosceles triangle are unlikely to be perpendicular. They must just mean the square of ... is equal to the product of the two distances ...
@Stan !!
Suppose we are given a fiber F of Gr(2,4)->P^2 which is the preimage of <V>, where V is the tangent vector of some 1-parameter group H of isoclinic rotations in 4D. Given any point x in S^3 (within R^4), H moves x around in some circle, which defines a plane, and that plane is an elt of F. We thus get a map S^3->F.
The set of all elts of S^3 which define a given plane (elt of F) under the action of H should be the intersection of S^3 with that plane, i.e. a copy of S^1. That means there is a fiber bundle S^1->S^3->F right?
18:08
@TedShifrin Ted! How are you?
Wow I am green today
You just wanted to match me, @Stan.
Hi @robjohn
Are you familiar with finite difference methods?
@evinda some
@anon: I'm now too preoccupied with everything that's going on here. But I think you'd better be very careful about orientations and non-orientations, as that's clearly rearing its ugly head here.
heya @robjohn
@TedShifrin yes that was my secret plan. Since i cannot hope to develop real math talent, I am forced to emulate Math SE avatars
18:10
You haven't kept me posted on your learning, @Stan.
Well, I had to postpone analysis til next quarter because of a back issue
I couldnt handle 4 classes
Too much sitting
I have written a code to approximate the solution of the heat equation. I want to consider uniform partitions in order to approximate the solution of the given boundary / initial value problem.

So we partition $[a,b]$ in $N_x$ subintervals with length $h=\frac{b-a}{N_x}$, where the points $x_i, i=1, \dots ,N_x+1$, are given by the formula $x_i=a+(i-1)h$, and so we have $a=x_1<x_2< \dots <x_{N_x}<x_{N_{x+1}}=b$ and respectively we partition $[0,T_f]$ in $N_t$ subintervals of length $\tau=\frac{T_f-t_0}{N_t}$ and the points are $t_n=t_0+(n-1)\tau, n=1, \dots ,N_t+1$, so we have $t_0=t_1<t_2<
@TedShifrin hey there
Meh, thats life.
18:11
Is that part of the health issue that made you move back to Chicago, @Stan?
@TedShifrin Guten Abend
Guten Abend, @evinda.
@TedShifrin yep. Its been a long standing thing. But its getting better without question. Just have to not push too hard. Patience.
@TedShifrin I am in reg econ now instead of honors
@TedShifrin the curve this quarter was supposed to be to a C
Which i dont really get why thats a good idea
Because now the regular econ classes are flooded with ppl who are too smart for the class
That's how we used to be, centuries ago, @Stan. Before we decided A was the average grade.
LOL did we. I missed the memo
18:14
what do you mean Ted?
Usually about 30% of my honors multivariable math kids bailed because they wanted an easy A. Many of them had more than enough talent but just didn't want to work.
I am gonna take economics 100 :D
Not working hard enough is bad.
for easy mark next semester
@evinda Why is $\tau=\frac{T_f-t_0}{N_t}$? shouldn't it be $\tau=\frac{T_f-0}{N_t}$?
18:15
hello
@TedShifrin the issue is there are too few students. There are like 15 and they are all amazingly talented. I am at the tail end of that. I dont mind the work. The tests are the issue. They are so hard.
@Sujith: Based on my translation of the problem, I do not find (in a simple case) that the curve is a circle. It seems to be a parabola.
@robjohn This formula is given. But in our case $t_0=0$.
@Stan: Usually professors say scary things because they want the students who want to learn, rather than get an easy grade. They usually don't grade nearly as harshly as the students think. Of course, if past years' records show that there are really are tons of Ds and Fs, then I'm wrong.
@Ted: Can I define a Hessian for any isolated zero of a vector field as a map $T_p M \to T_p M$? The answer is "yes" by passing to coordinates and writing it as $\nabla f$ - which I can do locally - but I don't want to do this
18:17
what about electives @TedShifrin
I mean ofcourse you wouldn't want to go to a hard elective
It's called the intrinsic derivative of a section of a vector bundle, @MikeM.
I took courses I was interested in learning things about, @Karim. I never chose on the basis of grades.
I needed the section to have an isolated zero to define this, yes?
I got a number of B's in electives. My world didn't crash to an end.
Um, I don't think so, @MikeM.
@evinda you can't make any estimates without more information
I need some help , I would like to prove that the cardinal of $HK$ for two subgroups of a finite groupp $G$ it's $\vert H\vert \vert K\vert/\vert(H\cap K)\vert$, I try using the orbit stabilizer formula. Where the direct product $H\times K$ act on the set $HK$ by $(h,k).x=hxk^{-1}.$
18:19
No metric needed.
@TedShifrin: You don't need a metric to define a Hessian? I guess maybe what I normally need it for is just to define the map $f \mapsto \nabla f$ as opposed to the other stuff.
@Stan, anyhow, with all your spare time, you can work more on my book :D
@TedShifrin should I take applied algebra or lie group ?
next semester
You need to know about manifolds to learn Lie groups.
I am more inclined towards lie group
18:20
@TedShifrin Yes, an excellent idea. It actually helped a lot. I felt much more comfortable with Hessians when they came up
we will learn about manifolds in this topology class
@Ted: Do you mind explaining how to define it or a reference? Google brings up nothing
@robjohn What further information? We are given initial and boundary conditions.. Or do we need something else?
Not differentiable manifolds, @Karim.
Hmm, @MikeM, really?
18:22
Interestingly, @MikeM, a lot of people are using that terminology for covariant derivative. Ugh.
A lot of elementary calc results, some stuff that looks possibly related but is unreadable, a lot of covariant derivs.
Yeah
So, I want to claim that if you have a vector bundle $E$ and a section $s$ with $s(p)=0$, then $ds_p$ is a well-defined linear map $T_pM\to E_p$. (In the case $E=TM$, you get a map $T_pM\to T_pM$.)
@robjohn Or can we pick any $N_x$ and $N_t$ we want?
Right. So you do need a zero, which is interesting
I think I answered an MSE question about this recently.
Yes, to check well-definedness, you absolutely need $s(p)=0$. Otherwise, a covariant derivative is needed.
18:25
what's the point? You just project something in $T_p E$ to $E_p$?
for the orbit we have Orb$(x)=\{(h,k).x: (h,k)\in HK\}$ so I need to focus on $hxk^{-1}$ to be an element of $HK$ as a set of products, but how can I continue?
No, such a projection would require a choice.
I would say there is only one orbit, $e$
because $hk=(h,k^{-1}).e$
@TedShifrin i feel like my understanding of the multiplier sucks. Like I can do Lagrangians, but I can't really explain what the multiplier does.my TA called it a "pentalty" of some sort, but I didnt really follow this
@Stan: That was one of the problems I had you do. The value of the multipliers gives you the instantaneous rate of change of optimal $f$ with budget $c$.
I'm not sure the TA knows what he's talking about.
18:28
@Ted OK... can you say how we get it, then?
@ted How can I find the function what gives the area of $AMNP$ depending only on x ?
@MikeM: I guess I just always did it locally and checked well-definedness independent of trivialization of the bundle because we're at a $0$ of the section, so the "non-tensorial" terms drop out.
:(
I'll just ask Ciprian about this case next week.
@Ramanewb: If you have the picture, you can figure it out. The point is that $M$ can be anywhere.
@ted Indeed, but I can't manage to find the value of $AP$ knowing $x$...
18:32
@TedShifrin bon soir Mr ted comment ça va aujourd'hui ?
@MikeM: OK, I guess you win. Since we're at a zero of the section, we get a canonical splitting $T_{(p,0)}E = T_pM\oplus E_p$, and then we just take the $E_p$ part of the derivative, yes.
@Ramanewb: How 'bout using similar triangles?
bonsoir, M @Agawa.
@TedShifrin: OK, thanks. I'll try and decode the case I'm in with that.
@ted I find
$PC = x$ so
$Area = x(14-x)$
@ted Is it true ?
18:40
Looks good to me.
@TedShifrin Suppose I have $f(x,y)$ s.t. $g(x,y) = c$. Why do the first order conditions give me $\nabla f = \lambda \nabla g$? In other words, I get how to calculate and solve a Lagrangian problem, but I don't really understand what a Lagrangian tells me about the relationship between $f$ and $g$. Why can a mere constant turn $\nabla g$ into $\nabla f$?
@ted I haven't used this theorem for ages although its so simple...
That why I couldn't think of it
It's the picture of the level curves (or surfaces) having to be tangent at a local extremum, @Stan.
@Ramanewb: It's important not to forget simple things :)
@ted Indeed
@TedShifrin o/ last day of holidays :(
18:42
@TedShifrin point taken , thank you , atleast now i can sleep thinking that its just a faulty question ,(but did you try googling a portion if the question) its like almost everywhere .
@AinzOoalGoal hey le pegase, you ll remain always un pegase pour moi
No, @Sujith. But I agree with Balarka. Horribly written stuff. They must mean some actual rectangle, but I have no idea what rectangle that is.
@Agawa001 Hippalectryons aren't pegases :o
@AinzOoalGoal: Qui ça?
Ooooh, it's Hippa.
Tout le monde prend des vacances?
le méchant hippa
18:43
Oui, surtout le méchant hippa.
@TedShifrin But how do I know such a tangency exists and will meet for a given function $g(x,y) = c$? For example, suppose $f(x,y) = \sqrt{xy}$ and $g(x,y) = 10x + 15y = 30$$. How do I know a point of tangency exists?
@hippa Hippa was on holidays, too...
Evidemment ...
@ted He's so lazy
@Stan: Existence of a max/min requires some sort of compactness arguments. One need not exist. But it if exists, the level sets must be tangent. Maybe go back to chapter 5 of my book?
18:45
I concur - source : am hippa
@Ramanewb: Et toi, t'es pas paresseux?
@ted Not at all, look, I'm working my maths !
But you've ignored my vector exercises for a month!
Aha I know which one it is :P he still hasn't done it ?
@TedShifrin ah! Interesting. So such a tangency need not exist. That explains why compactness is useful for this. Okay i will go and reread it. Bbl
18:47
@ted It's not a month but 2 weeks... And I couldn't do them because I was on holidays.
But I read them already and will do them soon...
Il y en avait deux, je crois, M le méchant :)
@TedShifrin oui its hippa chair et sang
@ted oui oui le deuxieme a l'air plus dur que le premier
@AinzOoalGoal Not quite...
18:50
J'en aurai beaucoup qui seront encore plus durs :)
@ted Wait till I've studied scalar product maybe
Im doing it soon I believe
Right. That's what I'm waiting for.
You don't need scalar prods for the first one
@AinzOoalGoal Not for the second one either.
And iirc you don't need them for the second one either
^
18:51
@AinzOoalGoal irc ?
OK, je m'en vais maintenant. A tout à l'heure.
Bye @ted
prends soins ted
@Ramanewbie iirc = if I recall
@AinzOoalGoal Ok
18:55
isnt internet relay chat ?
irc =/= iirc
19:24
Hello! Which abstract algebra textbook is the most terse one? I really need a terse one otherwise I will feel sleepy...
Hi @quid @anon @TobiasKildetoft !!!
Suppose that we have a function$ f(t_n+ \tau, x_i+h)$.
Can we find the taylor expansion for the above function only in respect to $x_i$?
is any finite group will have maximal subgroup ?
@Karim Zorn it.
19:33
yeah I guess so
boo my binary friends
@AinzOoalGoal pegases are more elegant than hippas
@Agawa001 heathen!!
zeus would prefere to fight riding a pegasus
Is anybody around that can help me with some Riemannian geometry?
19:48
@Memeozuki I don't know a thing about Riemannian geometry :c but in any case, just ask :-) hopefully someone else will know
Suppose I have a metric on an open set in Euclidean space that is conformal to the canonical metric and I have a straight Euclidean line that is the image of a geodesic in both metrics. In the canonical metric we see that the tangent field to this line and any parallel vector field along it all lie in the same 2-plane. I was wondering if in the conformal metric we also have the case that the tangent field and parallel vector field lie in the same 2-plane.
It makes intuitive sense to me but I'm trying to think of an argument avoiding christoffel symbols and high amounts of computation.
Let me know how that goes @Memeozuki
my isp sent me a warning telling me to stop fiddling with packets orelse you wouldnt receive any one from us
they suspended my internet flow for a whole day, what a shame, one mustnt learn some tcp protocoles
i didnt know how to contact em back so i wrote in my address bar "IT_IS_ETHICAL_YOU_AUTOCRAT_MOFOS.com"
i hope they wont suspend net traffic for another day
20:11
@Memeozuki: You're going to need to know how the LC connection changes when you change the metric by a conformal transformation. I don't know the formula for this, but certainly there is one. I would bet you'll have to do some computation but not absurdly much.
20:58
@evinda The values of $N_x$ and $N_t$ depend on how accurate you want your approximation to be.
r9m
r9m
@robjohn @Chris'ssistheartist I started a bounty here :-)
@AinzOoalGoal Is that avatar supposed to be skeletor? :P
@r9m lollllll, great!!! (+1) I bet the author has an elementary solution.
r9m
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@Chris'ssistheartist ah! don't loll me -_- .. I'm deeply frustrated about that one! :| I want that problem killed.Period.
@r9m not exactly :P
r9m
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@AinzOoalGoal kinda looks like Skeletor from The He-man show! :P
21:07
@AinzOoalGoal You're hippa, right?
r9m
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@Chris'ssistheartist Stoica must have had some crazy badass idea that simply eludes me!
@r9m lol, I think that you'll do my first 2 questions in AMM without pen and paper.
@r9m Well, he has very nice solutions.
@r9m After that, a hard time will come ... :-)
r9m
r9m
@Chris'ssistheartist :D okay
(just to have the rest of them accepted)
@BalarkaSen yeah
21:10
@AinzOoalGoal thought so. who else can have Ted's meme on his profile?
r9m
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@Chris'ssistheartist I sent that tan limit thingy I showed you long ago to AMM .. I'm not sure if it'd be selected :) Do they select problems coming from students?
@r9m I suppose YES.
@r9m If it's very hard, poor odds. One very nice proposal of mine was rejected.
r9m
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@Chris'ssistheartist :D 'Kay
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21:14
@Chris'ssistheartist well if they won't select it .. there are tons of other places I could send it to .. to get interested readers send nice solutions! :-)
@r9m This one is too nice to be true $$\int_0^{\pi/2} \int_0^{\pi/2} \frac{\log(1+\cos(x))-\log(1+\cos(y))}{\cos(x)-\cos(y)} \ dx \ dy$$ and it was rejected.
r9m
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@Chris'ssistheartist ow!! :) looks sinister! :)
@r9m I prepare a limit for AMM, this month, you will tell me later you never saw anything like that in terms of beauty.
I bet on everything!!!
r9m
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@Chris'ssistheartist :D oh yea! that's what I wanna see :D
@r9m You need recovery after you see it! Trust me! :-) I created thousands of problems, but that one is amazing!!!
21:17
@Chris'ssistheartist Is the book coming along nicely ? :D
@AinzOoalGoal oooooffffffffffffff ccccccccoooooouuuurrrrrrrrsssssssseeeeeeeeeee! :-)
r9m
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@Chris'ssistheartist kay kay!! don't kill me with foreplay alone :P (no offense)
@r9m :D
@r9m btw, do you see how to do the integral above? You can also try the squared version.
r9m
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@Chris'ssistheartist not off the bat .. I have to think :) .. you mean the square of the whole integrand?
@r9m OK OK
@r9m the integrand powered by 2.
r9m
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21:20
hai hai :)
hi
@BalarkaSen hi, do you know how can I prove prove that $\Bbb{Z}^n$ and $\Bbb{Z}^m$ are isomorphic iff $n=m$?
@user281591 you can prove that in general by replacing Z by any arbitrary commutative ring with unity.
Oh, you mean as groups?
@BalarkaSen this exercise it's in my course about groups yeah
actually, you can do the ring proof anyway, as you're looking at Z-modules.
@user281591 if those two were isomorphic, then so would be (Z/p)^n and (Z/p)^m
but that is impossible (count elements!)
21:33
@BalarkaSen Z/p=Z/pZ ?
r9m
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@Chris'ssistheartist AHA!!! I think I see! :D
@r9m Ah, but I remember now. I told you how to do that. :-)
r9m
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@Chris'ssistheartist wait! you did? :o I've forgotten! :O .. I made the change of variable $x+y = u$ and $x-y = v$ and used the fourier series of $\log \cos x$ ..
@user281591 Yes.
21:36
i don't see why you have this implication
@r9m Perhaps, all is possible. Maybe your brain kept the secret only (not the unimportant details like the source --- that's me). :D
@user281591 Think about it. (Z/p)^n = Z^n/pZ^n.
@r9m That's a different way. ;)
@BalarkaSen yep I need to think ^^
r9m
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@Chris'ssistheartist OMG! now I'm panicking! .. I really can't remember!
21:39
@r9m haha, well, if your way works, then you don't need mine. :-)
Hello
there
r9m
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@Chris'ssistheartist noooooo!! unfair .. please lemme know your approach too :) .. I don't think simply fourier will come to rescue if the integrand is squared
I am trying to proof that Kneser graphs $K(n,k)$ are connected if $n \geq 3k -1$
I was thinking about induction but I failed this way
Is there another possible simple way to prove this
@r9m It's a more general way I don't wanna talk about here. There is time for sharing stuff, no worry. ;)
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@Chris'ssistheartist okay!! :)
21:43
@r9m lol, you said nothing about the date of the problem on MO --- asked Jul 31 at 8:32
:D
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@Chris'ssistheartist I wouldn't have sent the solution to AMM if it was answered in MO (even assassins have a code of honor :P) .. but nothing was posted in MO that I already didn't know -_-
@r9m @TedShifrin said about me a couple of times I'm the most obsessed person with the calculation of integrals, series and limits he knows, but that's because he probably doesn't know you (well enough). :D
@r9m haha, OK :-)
rehi @TedShifrin
r9m
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@Chris'ssistheartist well my obsession was fueled by you mostly :P .. take responsibility will ya :P
@r9m LOLLLLL, very glad to hear that!!!! :-)))))
@TedShifrin look at that ^^^ :-))))))))
@r9m By the way, I saw you less talking about the creation process of problems. Don't you enjoy it?
It's far nicer to create than to solve, and I think you can learn much more this way!
r9m
r9m
21:52
@Chris'ssistheartist maybe mother-chan! :P but I am an aspiring assassin :P
@Memeozuki @MikeM: I would highly doubt that. Take the metric on $U\subset S^3$ coming from the usual metric on $S^3$, $U$ a disk. I'm finding it a bit hard to turn my spatial intuition into "conformally flat" intuition, but we could certain compute it.
I create at least 10 problems a day (I think) in this period, some are similar, but it's important the creation process, to push your creativity every day. @r9m
rehi @Balarka: Did you watch a fun movie?
Nope. I listened to an explanation of functor of points by someone over the commutative algebra room but most of it sounded Hebrew to me.
I ordered you to relax.
21:54
But listening to Grothendieck algebraic geometry is relax!!
:P
rolls 7 1/2 of 8 eyes
how did you roll 1/2 an eye? i could never do that.
r9m
r9m
@Chris'ssistheartist :) I rarely create problem .. I like to solve them :) (let the world laze around while I hog all the fun :P)
@Balarka: You have years of practice at interpreting those ill-formed geometry problems. What in the world was that problem asking?
@r9m Well, I also like to solve, sometimes in a good day I can even find 10 solutions for a problem (not that hard).
As the example with the logarithmic integral.
Sure, hard to believe, but I understand that.
21:57
@TedShifrin How'd I know? I skip ill-formed questions in my exams.
r9m
r9m
@Chris'ssistheartist I definitely wanna see all your solutions :D ..
Ah, no wonder they think you know no math.
@r9m I wanna add some to my book, not sure how many for the simple log variant, maybe 4-5.
I am not complaining.
I presume you find my exam questions well-enough posed :)
r9m
r9m
21:58
@Chris'ssistheartist okay .. for the book I wait then :)
BTW, that ladybug problem you hated was originally an exam question. They didn't get it ...
Yes, I did do your exam. It was good, standard, and encouraging :)
As I said, my homeworks are generally far, far, far more challenging (in part) than my exams. I think people who do that backwards are wrong.
@TedShifrin Something similar to that ladybug problem is that Chico-Groucho-whatever trio problem.
@r9m btw, for your tan problem, you have an elementary solution? That would be so amazing.
21:59
Yes, @Balarka, those are isomorphic.
Google Marx Brothers :P
Right.
Oh?

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