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2:00 PM
In the sense that you and PVAL were talking about it
 
To me when you say "rational thought", I get an entirely different perception than some enjoyable intellectual activity like music or math or whatever.
 
but the reason I try to "think rationally" about various things is for ethical reasons.
(in that I hope to adjust my actions based on this rational thought).
 
need irrationality necessarily contradict with ethical notions? I would say no.
(was this Daminark's question too?)
 
If they're not identical, they're adjoint
Peter May intensifies
 
2:04 PM
0
Q: A nonlinear ODE that came up while finding Eigen function for a p-Laplacian without any boundary condition

Rajesh Dachiraju$$r^{d-1} f(r)|f(r)|^{d-1} - A\frac{d}{dr} \left( r^{d-1} f'(r) |f'(r)|^{d-1}\right) = 0$$ solve for function $f:\mathbb{R}\to\mathbb{R}$. $d \in \mathbb{N}$, $A \in \mathbb{R}$, $A,d$ are constant. Has there been any work on this equation? I tried Wolfram but its not able to, in the computati...

 
By ethics I mean philosophy in order to answer the question "how to act"
 
its an ode problem
 
So I think rationally about a lot of things because I think it helps me figure out "how to act"
 
Oh by the way on prodding from Dylan we did do a couple computations here and there. The first class we had was the bonus, which was basically just review. The second one was a kinda random QA session, in which stuff happened
Realized that $K(\mathbb{Z},1) = S^1$
 
What is K(Z, 2)?
something to think about
 
2:08 PM
Actually $K(\mathbb{Z},1) \simeq S^1$
@BalarkaSen BS^1 obviously.
 
Actually wait I just remembered that left and right don't work the way I want them to, so $H^1(X;\mathbb{Z}) \ne \pi_1(X)$...
 
$H^1(X; \Bbb Z)$ is actually the abelianization of $\pi_1(X)$
 
@BalarkaSen no
 
Sorry.
Dumb
I meant to say $H_1$
 
2:11 PM
It's $\Hom (\pi_1, \Bbb Z)$.
 
Right, that.
apparently I have exterminated all my rational thoughts, or at least a good part of it
 
So we want to find something whose loop space is $S^1$
Also @Balarka : Rel8able content
 
initially i thought it was a rickroll
but it's worse
 
Honestly though
 
how to write intersection in \something
 
l m a o
 
i want to find the volume of $S \ ^ 2 \cap \{(x,y,z) : z \ge \sqrt{x \ ^ 2 + y \ ^ 2} \}$
 
the subtitles are gold
 
so in spherical coordinates we have $\phi \in [0, \pi/4] , \theta \in [0,2\pi] $ and now im looking for the boundaries of $r$
someone can help with that?
 
Is $S^2$ supposed to be $B^3$?
 
2:19 PM
yes sorry
 
Good morning, chat!
I actually slept in normal sleeping hours.
 
I'd imagine that works out reasonably nicely in Euclidean coordinates.
 
@Fargle :O
 
maybe, but could you help with finding the boundaries for $r $ ?
 
I have started doing that
maybe i should break it
 
2:20 PM
Don't.
 
@Semiclassical maybe you can :P
 
Write your inequality in terms of spherical coordinates
 
I've done nothing but feel like crap the past week or so because of my sleep schedule.
 
and solve for $r$
 
@PVAL-inactive what do you mean?
 
2:21 PM
Write z \ge x^2 +y^2
 
i have $r \ ^ 2 sin(\phi) = r \ ^ 2 cos(\phi)$
 
in spherical coordinates
 
it is $z \ ^ 2$
 
It's illuminating to take a special case. Suppose x=y=0. What's the bound on z in that case?
 
@Daminark I don't know why but I am listening to this right now.
 
2:22 PM
Hello People! I have been trying to solve a pde but stuck in a particular place. Can anyone have a look at it? math.stackexchange.com/questions/2349183/…
 
@Semiclassical $ x \ ^ 2 + y \ ^ 2 \le z \ ^ 2 \le 1$
 
@Semiclassical he wants the bounds on r in terms of \phi and \theta
 
Sounds dark @Balarka
 
Sure, I'm making a specific point
 
but this make no sense if we plug $x=y=0$ we get $z\in [0,1]$
@Semiclassical hm?
 
2:26 PM
@Daminark It's just about poisoning pigeons
A fairly innocent hobby
 
So that means that, if we're restricted to the positive z axis, the range of possible distances from the origin are 0 to 1
So $0\leq r \leq 1$ in that case.
 
"Innocent"
 
how this make sense?
 
Any topologists here?
 
@Semi what he needs to show is that that range works for a fixed $\theta$ ,$\phi$.
 
2:28 PM
the cone cuts the sphere in some height and we want the volume of what above that height
 
@Daminark Here is another good one.
 
can someone please explain to me this compactness argument?
 
@Liad since your equation puts no condition on $r$ and since you are inside $B^3$ you are looking at $0<r<1$.
 
sure. Where I was going is that the upper bound on r is coming from the sphere condition
 
I'm having trouble explicitly constructing the covering by rectangles
 
2:29 PM
pde experts? anyone?
 
this is Hatcher's proof of Van Kampen's theorem
 
@Semiclassical the upper bound of $r$ is always $1 $ doesn't it? i thought this is obvious
 
@Semi If you do the case he wrote originally with z replaced with z^2, you will get a real restriction on r coming from the other equations
namely r> cos \theta
 
@PVAL-inactive we want points above the cone so there are condition on $r$, or am i missing something?
 
sure. I wasn't disputing that
 
2:31 PM
i am a bit confused :/
 
@Liad z^2=x^2+y^2 isn't a cone.
 
But I think whatever point I intended is at best a distraction at this point
Um, yes it is
 
@PVAL-inactive it is a cone :P
 
Take the cross-section with y=0
 
@wilkersmon The existence of that is a consequence of the Lebesgue covering lemma.
 
2:32 PM
oh im dumb
 
@BalarkaSen thank you, I'll look into that
 
but yeah essentially it's pretty clear that if you are on a spherical angle on one side of the cone
you get every radius you want
 
$f^{-1}(A_\alpha)$ cover $I \times I$; choose a partition of the $I$ factors so that the subrectangles all fit inside the cover
 
so $r \le 1$ is clear to me
but how do we get the lower bound?
 
@Balarka noice
 
2:34 PM
we need to find where the line that takes a point in the sphere to the orgin intersects the cone, right?
 
The line from that any point in your region to the origin.
 
Hi Balarka
 
and that line intersects the cone only at the origin.
 
huh ?
 
2:35 PM
Have you ever came across the ODE I mentioned?
 
how ?
 
That's what varying r looks like.
 
I am not a ODE person
 
You take the line from the point to the origin and move along that line.
 
2:36 PM
doesn't this region looks like a ball that we cut in height (in z-axis)$a$ for some $a$
 
@Daminark that tune tho
 
so that's my mistake. but i dont see why :/
 
Also I'm a bit stumped on $K(\mathbb{Z},2)$
 
you take a ball and intersect it with a cone
 
2:37 PM
@Liad I thought you were responding to me with that and I was just like heh?
But yeah so we want to find a space whose loop space is $S^1$
 
@Daminark hehe :P
 
When is the fundamental group of a space homotopy equivalent to its raw mapping space?
 
@Daminark What's an easy space with $\pi_2 \cong \Bbb Z$?
 
$S^2$
 
@Liad both the ball and cone are centered at the origin
 
2:39 PM
alright , i agree with that
 
So you get kind of a cone looking thing with a curved bottom.
or curved top.
 
what's the easiest reason it's not a $K(\Bbb Z, 2)$?
 
huh
i see the mistake now
 
hi chat (from laptop now)
 
yea $r\in [0,1]$
@PVAL-inactive thanks
 
2:40 PM
Still this can all be done by noticing that the equation doesn't put any relation on r
 
Hopf map is not nullhomotopic
 
@Liad if you are careful
 
to put it in slogan terms: The intersection of a sphere with the upper part of a cone, both centered at the origin, is a spherical cone
 
@Daminark Right, $\pi_3(S^2)$ is not zero. What if I take a 4-ball, and glue it's boundary 3-sphere to $S^2$ by the Hopf map?
 
and this is somewhat essential because you'll probably be given examples which aren't so easy to visualize.
 
2:41 PM
That is to say, take the mapping cone of the Hopf map $f : S^3 \to S^2$
 
so we get $\phi \in [0,\pi/4] , \theta \in [0,2\pi ] , r\in [0,1] $ and it is easy from here. thank you both @PVAL-inactive @Semiclassical
 
$K(\Bbb Z, 2)$ is a group and $S^2$ cannot even be an H-space by cohomological arguments.
 
Right. It's not bad once you work it out.
@MikeMiller I momentarily misread that as "by cosmological arguments"
 
though to be honest I think the the Euclidean way is just trig sub at most.
 
The very universe objects to your mathematics!
 
2:43 PM
@MikeMiller Do you know it's a group off the book? $K(G, n)$ is at best an H-space with homotopy-associative product and has identity and inverse upto homotopy, right?
 
I don't think I know any volumes which are easy in spherical coords. , but out of the scope a good calc. student could do in Euclidean coords.
@Balarka I think it's the unitary group of a Hilbert space
 
Yeah. I mean, a spherical cone is trivial enough in spherical coordinates, but
it's not that hard in cylindrical coordinates either.
And even Cartesian coordinates isn't impossible, though it's certainly less practical.
 
@PVAL Ah, I guess you could say that.
 
Though, come to think of it
Wouldn't the trig substitutions amount to spherical/cylindrical coordinates in disguise?
 
It's really not far from the "standard description" of K(Z, 2) though
(which daminark is supposed to figure out so i am not saying it)
 
2:47 PM
or is the spherical coordinates just trig sub in disguise?
 
pretty much.
 
in cylindrical coordinates we have $z\in [0,1] , \theta \in [0,2\pi] $ and how we get $r$ here?
huh wait
 
@Balarka actually what I said is contractible.
 
well, if you think about it in cylindrical coordinates, there's two relevant boundaries.
 
@PVAL so you quotient it with S^1?
by
 
2:51 PM
if you're close to the top, the boundary in r (distance from the z-axis) is a circular arc.
if you're not, then it's a straight line (part of the cone)
so you'd need to split $z$ into two integrals.
 
I didn't understand Neves's construction of the Hopf map too well but I dunno, this whole toying with spheres things sounds either toral or $\mathbb{RP}$ type
 
Alternatively, you can think of $r$ going from 0 to some max value, and saying that $z$ is bounded above/below appropriately.
 
wait
 
Either way it's not as neat as in spherical coordinates.
 
yea but i want to find the boundaries for practice
 
2:53 PM
@Daminark I just wanted to know what $\pi_3(Cf)$ is but yeah your guess is close
 
Sure.
 
$\theta \in [0,2\pi]$ either way
 
you don't need to know much about the Hopf map to compute that thing
 
So, if I look at the spherical cone, what's the largest possible value of $r$ you could have?
 
1
huh wait
 
2:53 PM
I mean, all I can say about the Hopf map is that it isn't nullhomotopic
 
Spherical $r$, or cylindrical $r$?
 
yea this was my mistake in replying fast
 
(This is why people use $\rho$ for spherical radial coordinate. But I hate that---rho is already density in physics, so it conflicts with one of the main applications of multivariable calculus. I like $r$ for spherical and $s$ for cylindrical, but that's a matter of personal taste. I know that doesn't sway math people, though...)
 
that i meant $r$ in spherical .
 
@Daminark how can you say that
 
2:55 PM
so $x \ ^ 2 + y\ ^ 2 + z \ ^ 2 \le 1 $ --> $r \ ^ 2 + z \ ^ 2 \le 1$
 
There's a cute argument with the HLP
 
@liad right.
 
i like the H.I.
/L.N.
Hi @Ted
 
Oh hell, Semiclassic is complaining about $\rho$, yet again!
Hi, Balarka'.
 
shakes fist
 
2:56 PM
hehehe
 
Heya @PVAL. BTW, I fixed that stupid error, but as far as I know the OP never responded.
 
I mean, the better way to put that is that this is all I recall about it
 
@TedShifrin They rarely do
 
Hi @Demonark
 
Seriously, one of the main reasons to do multivariable calculus is for physics/engineering computations. And one of the main reasons to do so is to compute things involving charge or mass distributions.
 
2:57 PM
Well, also that the preimage of every point was a circle I think?
 
I did however just have someone respond to like 5 of my comments in quick succession
 
@Semiclassical so $z \le \sqrt{1 - r \ ^ 2}$
 
Actually, I have interacted with him before. He's a faculty member somewhere on the east coast (Virginia?).
 
which was quite nice.
 
Hey @Ted!
 
2:57 PM
@liad Right. Or alternatively $r\leq \sqrt{1-z^2}$.
 
@BalarkaSen You can make K(G,n) a group on the nose reasonably easily. It's Omega K(G, n+1), and you can model the loop space with Moore paths [0,s] -> K(G,n+1) that gives it a product associative on the nose. I guess I still need to do inverses but I'm sure there's some similarly stupid trick.
 
Greetings, @Liad. I thought you already took your exam?
G'night, @MikeM.
 
Either one of those will work, but not at the same time.
 
@TedShifrin Sunday :P
 
Oh.
 
2:58 PM
@MikeMiller Ohhh. I see.
 
You either allow $r$ to vary freely, and constraint $z$ thusly, or vice versa
 
That's really cute
 
@Semiclassical yea so $ z\ in [ 0, \sqrt{1- r \ ^ 2 } ] $ and we are done?
 
@Mike for some reason I can't isolate, "group on the nose" reminds me of the justification we got for cofibers giving you the exact sequence of pointed sets
 
That's why EM spaces have such good properties in some senses. They're loop spaces to arbitrary order.
 
2:59 PM
"Hang the homotopy on the cone!"
 
If you're writing it like that? Nope.
 
(I'm sure PM would agree with me here.)
 
right we need the boundaries for $r $ :P
 
on the nose means it's happening actually actually
 
Well, that's true. But there's another issue: $z\leq \sqrt{1-r^2}$ is the boundary from the sphere.
 
2:59 PM
as opposed to happening upto homotopy
 
You've also got the cone.
 
@Daminark COMPUTE $\pi_3(Cf)$
if not visually, do the algebra hrr drr
 
Here's a historical question I find myself wondering about, in retrospect.
Mathematicians using $\rho$ for spherical radical coordinate, or physicists using $\rho$ for density.
 
@Semiclassical $z \in [r , \sqrt{1 - r \ ^ 2 } ]$
 
Right.
 
3:01 PM
what about $r$
 
Well, what's the largest $r$ for which the above interval makes sense?
 
i need to tell how much to go in the $[xy]$ plane but im not sure
 
mumbles Draw Pictures
 
ahh , i must get better at this. the question that do not need to find this regions im ok with , but this questions that we need to find the boundaries for the coordinates im not sure sometimes :/
 
@Mike @Dami okay here's a super bare bones argument proving the Hopf map is non-trivial.
First consider the map $S^1$ disjoint union $S^1 \to S^2$ which maps to two points
actually nevermind this doesn't
work
isn't there a way to do it just with HLP
 
3:06 PM
what is HLP
oh hmotopy lifting
 
i cant find the restrictions of $r$ @Semiclassical @TedShifrin
 
i dunno, i don't think
 
Say, population can be modeled by $p(t)=p(0)e^{kt}$, right? Are the units for $k$ given by 1/time? If not, then how does $e^{\text{sec}}$ comes out, as far as units.
 
unless you mean, like, proving the boundary map in the long exact sequence is an isomorphism
 
@Liad: I came in late. What is the actual region?
 
3:07 PM
Maybe it'd help to think of the Hopf map as mapping into $\mathbb{C}$ or something?
 
@TedShifrin in the ball , above the cone
 
That same question as the other day in spherical coordinates?
 
@TedShifrin already did it in spherical coordinates
 
So where is the region the fattest? That will tell you what the projection in the xy-plane is.
 
yea i wanted to try using cylindrical coordinates
 
3:08 PM
@PVAL Yah. You lift the null-homotopy of the Hopf map to S^3. But then you need to know pi_3 S^3.
 
@Balarka @PVAL @MikeM Demonark ... If the map were nullhomotopic, the fibers would have linking number 0. ?
 
I guess you just need to know S^3 is not contracfivle
 
anyone know about eigen functions of p-Laplacian
 
@TedShifrin We all know this argument, I think. PVAL is thinking of a different argument
 
@TedShifrin but dont we need to do it for each hieght $z$ ?
 
3:09 PM
right
 
@Ted Sorry, to clarify, none of us are asking for arguments. Daminark is learning this for the first time and I was asking if he actually had an argument or was going off memory.
 
I'm sorry for interrupting ... :)
 
@TedShifrin i mean , $r$ changes when we go up in the cone / ball
 
@Liad: You should ordinarily set up cylindrical coord integrals in the order $dz\, dr \, d\theta$.
 
@Mike That was definitely the argument I had in mind.
There's also the cute construction piggy-backing on that
 
3:10 PM
So find the region it projects onto in the $xy$-plane. Then, for such a fixed point, ask "what does $z$ do?"
 
@TedShifrin right and i already did find the boundaries for $z $
 
So what points in the $xy$-plane does this region project down onto? Look where it is fattest.
 
of the map $T^3 \to S^3 \to S^2$ which is zero on all homotopy and homology groups but not null homotopic.
 
(You should look at some of my videos where I do these problems, Liad ...)
 
yea it is the fattest when $\phi = \pi /4 $
 
3:12 PM
The region looks like an ice cream cone with ice cream on top. What is the biggest radius it ever has?
 
The point being a homotopy lifting a nullhomotopy would have to live inside a fiber of the hopf map
so the degree 1-map from the torus to S^3 would be null-homotopic.
 
@TedShifrin when you are at $z \ ^ 2 = x \ ^ 2 + y \ ^ 2 $ , on the sphere
 
ok, imma go get a little nap
 
So use algebra to find out how fat it is there. The projection is going to be a disk of what radius centered at the origin?
 
too tired
 
3:14 PM
Night, Balarka'.
(@Liad: Suggestion — write both equations in cylindrical coordinates to start with.)
 
$ r\ in [0 , \dfrac{1}{\sqrt{2}} ]$ @TedShifrin
 
Oh wait a second, checking back through notes we never actually proved the Hopf map stuff, he just said it and then proved that higher homotopy groups of $S^1$ were trivial
 
@PVAL-inactive Yah, it's a nice one.
 
@Liad: Are you sure?
 
$z \ ^ 2 = r \ ^ 2 , r \ ^ 2 + z \ ^ 2 = 1$
 
3:16 PM
Aren't the equations $z=r$ and $r^2+z^2-2z = 0$?
@Liad: Wait. We were doing a sphere that was shifted up last time we talked.
 
huh, now we dont :P
 
GRR.
 
:)))
 
OK, then $0\le r\le 1/\sqrt2$. Done.
 
i kind of starting to like it
 
3:18 PM
@TedShifrin That seems like a hard thing for a calc student, but you're the expert I guess.
 
ran into a former student of mine and got distracted, woops
 
@PVAL: I grrred because he changed problems on me.
Last time we talked he had a shifted sphere and we discussed that :)
 
I was trying to make a pun on (GRR)
 
The point I was going ot make earlier is that, if you've got $z\in [r,\sqrt{1-r^2}]$, then that interval only makes sense when $r\leq \sqrt{1-r^2}$. @liad
 
@PVAL-inactive boo.
 
3:19 PM
so you need $r^2\leq 1-r^2\implies 2r^2\leq 1\implies r\leq 1/\sqrt{2}$.
 
Good grief, @PVAL. Filter your remarks.
@Semiclassic: We finished.
 
Yeah, but I figured I should follow up on what I was saying before I got distracted.
 
Not until after Liad "forgot" to tell me he'd changed problems from the last time we talked about it.
 
next time to go in the middle of the party ;-)
 
Yeah, I noticed
 
3:20 PM
glares at Liad
 
yea i should tell you when i change problems :P
 
I had entirely forgotten the shifted sphere problem, so I was a bit lucky in that regard I guess :)
You could also set up the bounds like this.
 
I'm old and forgetful, but not so forgetful that I lose my limousine.
 
any $\theta\in[0,2\pi)$, and any $z\in[0,1]$, but $0\leq r\leq z$ when $z\leq 1/\sqrt{2}$ and $0\leq r \leq \sqrt{1-z^2}$ when $z\geq 1/\sqrt{2}$.
That works, but um
Between the two, I'll take the other.
 
3:35 PM
Hmm. I'm embarrassed to admit: I'm looking at a PDE problem on the main site, and I can't for the life of me remember when the separation constant is allowed to be positive vs. negative vs. zero.
My recollection was "allow any of them and let boundary conditions sort them out."
nm, I think I figured it out.
 
3:50 PM
Analysis will prevail
Also, hi chat
 
(The funny thing is that the answer to the PDE is literally just a linear function of $z$. No complicated Bessel stuff or anything.)
 
Any cool maths today ?
 
The alt text for today's XKCD makes me chuckle
Not sure it counts as cool math, but here's something I'm thinking about @astyx
 
I'm all ears
 
Let $f(x)=xe^x$.
And suppose, for whatever reason, I want to take the Legendre transform of that.
 

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