Mathematics - 2016-02-12 (page 2 of 3)

Akiva Weinberger

13:02

@TedShifrin Oh, you were cutting it in half the other way? I thought you were cutting the donut in half the way one would actually cut a donut in half (i.e. leaving an annular surface on which to spread cream cheese).

Tobias Kildetoft

@AkivaWeinberger Why would you put cream cheese on a donut?

Akiva Weinberger

I mean bagel.

Tobias Kildetoft

Ahh, that sounds a lot better

Akiva Weinberger

Same shape. They were actually just talking about a torus.

Though, now that I think about it, cream cheese on a doughnut probably isn't so bad…

skull petrol

Cream cheese on a sweet donut?

:-/

Idle001

13:06

@skullpetrol wb pal :D

Akiva Weinberger

Can't know until you've tried it

skull petrol

Thanks pal @Idle001 :D

Akiva Weinberger

Like pineapple on a pizza. *ducks*

Idle001

hmm yes anon and alec were been long term absent

skull petrol

yup

Idle001

13:07

i used to see their icons permanently

skull petrol

Alex too.

"That last LinkedIn request set a new record for the most energetic physical event ever observed. Maybe we should respond." "Nah."

Secret

14:03

Agreed. Even if we attempt to push further by trying to enumerate all the possible ways the proof can be subverted, or at least the existence problem on whether it is possible given a truth statement, it's basically just pushing the same question further back and not really did much

hmm...

bruce.banner

Hi All

N3buchadnezzar

heya

bruce.banner

I have a question about the Flajolet–Martin algorithm, aka Probabilistic Counting - https://en.wikipedia.org/wiki/Flajolet%E2%80%93Martin_algorithm. I am trying to draw understand the distribution.
Excuse the rather basic quesiton - my theory needs a lot of work.

algo.inria.fr/flajolet/Publications/DuFl03-LNCS.pdf and the LogLog paper.

"n fact, in probabilistic terms,
the quantity
R
is precisely distributed in the same way as 1 plus the maximum
of
n
independent geometric variables of parameter
1
2
. "

So the binary of the hash is geom, ok...

question is, why not just use p=1/32 instead of separate geom rand vars

Idle001

15:07

hello @I'manartist see someone gave a shot to ur problem

I'm an artist

@Idle001 yeah, still my problems are pretty advanced (both for MSE and MO).

Idle001

@I'manartist ofcourse, cant dare to get arround em

I'm an artist

@Idle001 hehe, I was just in the middle of some awesome research. :D

Any ping reminds me to get a break from working, and maybe trying to east something.

It's 17:11 here and didn't eat anything. My mind is sharper when I eat less.

Idle001

ew, i just bear a cup of coffee in my hand

I'm an artist

@Idle001 My work is more important than me (it may sounds a bit of nonsense, but well, it is so).

GPhys

15:14

The following exercise from my book seems extremely esoteric to me. Is there a motivation for this?

Idle001

@I'manartist that is what militants say, my case is more important than me, or my weapon is more important than i

I'm an artist

:D

Idle001

anyone who has a case to defend, or a goal to reach

bbl

need another cup of coffee

Albas

arxiv.org/abs/1409.7991 Hehe ,Nice read

Semiclassical

@GPhys: that looks to be an example of an integral transform with kernel $k(x,y)$ supported on $[0,1]$.

Balarka Sen

15:21

@GPhys I wonder if that's an isometry.

GPhys

it's from an open source book, if you wanted to browser the chapter: jirka.org/ra/realanal.pdf

Semiclassical

so the problem amounts to showing that, if you change your initial function 'smoothly', then the transform changes smoothly as well

integral transforms come up a ton in applications, so it's not as esoteric as it seems

Mike Miller

@GPhys: This is the integral form of a differential operator, which makes it easier to allow non-smooth input. $k$ stands for kernel.

If $k$ is nice enough such an operator actually increases the smoothness of your input.

Semiclassical

i'm more used to examples with the kernel being supported on the entire real line or half the real line

but that's just a change of variables

Mike Miller

Yeah me too.

But they didn't want to talk about spaces of functions that weren't continuous and the easiest way to get a Hilbert space is to restrict to ones on the interval.

Semiclassical

15:27

say, by substituting $y=\tanh u$

@MikeMiller good point

i can actually come up with a physics example where that'd come up, now that i think of it

take $y$ to be the position coordinate of a quantum-mechanical particle in a box, and suppose $f(y)$ is its wavefunction in position space

then the choice of $k(x,y)=e^{i x y}$ gives (modulo some constants/rescalings) the momentum-space wavefunction as $\phi_f(x)$

GPhys

Ah, thanks all. It seemed fairly "out there" relative to the other exercises.

Mike Miller

Integral operators are pretty important I suppose.

Semiclassical

yeah. and integral transforms are huge in engineering/physics/applied math

albeit usually in the sense of "what does this particular kernel give us" rather than kernels in the abstract

my example isn't actually quite right for the problem as stated, since the wavefunction are generically complex-valued (and the Fourier kernel $e^{i x y}$ definitely is)

probably one would need to head over to the regular wave equation (vibrations of a string etc) to find a pure example

Balarka Sen

Hi @iwriteonbananas.

$f(x, y) = 2x^4 - 3x^2y + y^2$ is a weird beast. The origin is a critical point which is a local minimum restricted to any line through origin, but is not a local minimum point for $f$.

What's more, apparently the Hessian of $f$ at $(0, 0)$ is degenerate. So annoying.

Mike Miller

@Semiclassical: I mean usually we're considering specific PDE too, so it's not much different.

Semiclassical

15:42

true

Mike Miller

The cases where we're not amount to the general theory of elliptic linear PDE

Because what else can we do without restricting ;P

N3buchadnezzar

Oh, hi

Balarka Sen

Actually, I wonder if it's possible that critical pt of $f$ is the origin which is a local minimum for $f$ restricted to any line through origin but is not a local minimum for $f$ - yet hessian of $f$ is nondegenerate. I doubt this.

Right, of course it's not possible.

N3buchadnezzar

Anyone have any experience with optimization? Thinking about asking a question on main, but unsure how to formulate it.

GPhys

I'll have to try that problem later. It will involve going back and checking more definitions than usual for this chapter.

Balarka Sen

15:45

(argument: if hessian is nondegenerate, then it must locally look like either $x^2 + y^2$, $x^2 - y^2$ or $-x^2 - y^2$. So it's either a local minimum, maximum or a saddle. the first two cases are outright cancelled off. it cannot be a saddle, because then you have a line along which $f$ increases and a line along which $f$ decreases, so the condition is not satisfied)

@N3buchadnezzar Hi.

iwriteonbananas

@BalarkaSen hey man

N3buchadnezzar

I have some ovens, some water taps which produces a different amount of water / heat etc. Say for simplicity sake they are most efficient when they produce at a given rate.
So say $x_1 = 10$, $x_2 = 15$ and $x_3 = 30$. The efficiency is zero of the oven is off, otherwise it is given as 100*(x - a)^2.

Balarka Sen

Hello @iwriteonbananas.

I'm an artist

Is any done with this beautiful one?

N3buchadnezzar

My problem was to figure out what rate to run the ovens / taps to maximize the efficiency when you want to produce T heat. Seems complicated

Semiclassical

15:52

i think that if you're waiting for a "no pencil and paper" response to an integral that involves Ei(x) in a nontrivial way, you'll be waiting a while

iwriteonbananas

@BalarkaSen I'm going to be algebraic topology 1 teaching assistant next semester. I'm scared.

Semiclassical

...though seeing as it numerically seems to come out to $\pi^3/4$, maybe i'm being hasty

Balarka Sen

@iwriteonbananas Congrats! What does alg top 1 entail?

Mike Miller

@iwriteonbananas: I will also be TAing algebraic topology I at Muenster.

iwriteonbananas

@BalarkaSen basic category theory notions, Van Kampen, singular and cellular homology, and applications. Essentially the same stuff that I did last semester.

@MikeMiller No way! You're going to Muenster?

Mike Miller

16:05

Oh, no, I thought that's where you were.

iwriteonbananas

Hehe, no but it's not too far off.

Balarka Sen

@iwriteonbananas Cool.

iwriteonbananas

@BalarkaSen I don't know if it's cool, i'm kind of shitting my pants

Semiclassical

When does the next semester start?

Balarka Sen

you'll be fine. You know your stuff.

iwriteonbananas

16:07

@Semiclassical Mid april.

Semiclassical

so about two months to get prepared

Balarka Sen

huh, apparently I haven't forgotten how to extremize.

Brennan.Tobias

hi

is anyone interested in number theory/prime gaps

if so you could look over an answer to a question I asked?

Semiclassical

@BalarkaSen one thing worth noting is that that function factors as $(x^2 - y) (2 x^2 - y)$

which means that if you focus on, say, $x\geq 0$ and substitute $z=x^2$ (which is increasing on this domain) you get $(z-y)(2z-y)$ which is much nicer to analyze

Balarka Sen

@Semiclassical Yes, so $0$ is attained all along $y = x^2$. Indeed, you choose $y = x^2 + \epsilon$ to get $-\epsilon(x^2 - \epsilon)$ and make sure $\epsilon < x^2$ to guarantee a value smaller than $0$.

Semiclassical

16:17

right.

Balarka Sen

So even if $(0, 0)$ is local minimum restricted to any line, you can get you value to be less than $f(0, 0) = 0$.

So it's not a global minimum.

Semiclassical

yeah

Balarka Sen

Idea: I should badger you with my calculus problems whenever Ted's not here (i.e., majority of the times). Let me know what you think :)

Semiclassical

haha

not sure he'd like that :P

Balarka Sen

whom?

Semiclassical

16:19

Ted

you're the person who's supposed to be learning it, after all

returning to the question, though: in $(z,y)$ space you also have an explicit saddle point at the origin, which means there's a line going through the origin which has the fastest descent

and this in turn maps back to $(x,y)$ space to give a parabola of fastest descent

Balarka Sen

Oh, I mean, I am not going to ask you to do my problems - that'd miss the point of me learning calculus. But I was merely wondering if you'd like to hear a few of my own understanding and realizations.

Also, I come up with silly questions occasionally. A nudge towards the right direction can help me catch my silliness.

Semiclassical

yeah, i understand. i was just being a pest :)

though maybe the better word is 'gadfly', which is what it sounds like you need

Balarka Sen

haha

Clarinetist

Anyone here know of a quicker way to do this problem?

0

Q: Predictive Distribution with Normal Prior

Given $\Theta = \theta$, let $X_1, X_2, \dots, X_n, X_{n+1} \sim \mathcal{N}(\theta, \sigma^2_0)$ be independent. $\Theta \sim \mathcal{N}(\theta_0, \tau^2_0)$. What is the easiest way to find the distribution of $X_{n+1}\mid X_1, \dots, X_n$? I am looking only for hints. How I would start thi...

Balarka Sen

i am sure Ted'll be very glad if I badger him less.

:P

Semiclassical

16:25

you may have a point there

Balarka Sen

@Semiclassical What do you mean by the fastest descent there?

Semiclassical

actually, i think i misspoke back there

so i need to check something first

Idle001

i called this week , the week of combinatorics, since my main interest to answer combinatoric questions

I'm an artist

17:02

I'm done with my work for half an hour.

Semiclassical

hey @pval

PVAL

@Semiclassical hi

Semiclassical

i realized one nice thing about your matrices, which makes producing them actually very simple in mathematica

first, if $A$ is your lower triangular matrix of ones, the inverse is lower bidiagonal with +1 on the main diagonal and -1 on the subdiagonal

which amounts to $A^{-1} = I-S$ where $S$ is a shift operator

PVAL

@Semiclassical I interpreted the signature another way, and the whole problem fell apart kind of easily.

Semiclassical

ahhh

for general sizes, or for your specific cases?

PVAL

17:14

At least for n sufficiently large (which is larger than I computed).

I haven't thought about when $|sig|/dim<2/3$ in general but I imagine a similar argument works as long as the blocks and number of blocks is large enough.

Semiclassical

probably

anyways, what i was going to point out is that the total matrix is the symmetrization of a certain bidiagonal block matrix, whose elements are $A$ on the subdiagonal and $-A$ on the main diagonal

which i can write as a Kronecker product of $A_p$ with $-I_q+S_q$ where $S$ is the shift operator i said above

which, using the characterization i gave above, means that the total matrix is the symmetrization of $-A_p \otimes A_q^-1$

and that's probably the easiest for mathematica to define the matrices

Idle001

hmmm

PVAL

@Semiclassical Somewhat amazingly the signature for r blocks of size q is $\sum_{i=1}^{q}\sum_{j=1}^{r}sign(\sin\pi(\frac i{q+1}+\frac j{r+1}+\frac12))$. Which is a lot easier to deal with than the forms themselves.

Semiclassical

17:30

hmm!

now i want to prove that :P

not sure why you'd do $\sin(\pi/2+\Theta)$ instead of $-\cos\Theta$, though

PVAL

You can replace the 1/2 with a more general formula involving a 3rd summand, for related forms.

Semiclassical

ahh

what kind of related forms?

PVAL

In theory I think I know how to write the matrices corresponding to these forms but I havent went through it yet.

Semiclassical

mmkay

Ted Shifrin

@Akiva = DogAteMy ... We were cutting the bagel horizontally (the way you would spread cream cheese). I stand by my remarks :P

hi @PVAL @Semiclassic @Idle

Semiclassical

17:42

hi @ted

Mike Miller

You should cut the bagel along a trefoil

PVAL

hi

Mike Miller

PVAL

You should give me the bagel so I can eat it.

Ted Shifrin

@MikeM: Actually, Karim needs to see the picture cutting it along a 1,1-curve. But I told him to give me a CW structure if we slice it horizontally.

Mike Miller

17:44

@PVAL: If someone gave me a bagel cut along a trefoil I would eat it.

Balarka Sen

Hi @TedShifrin.

Ted Shifrin

Hello @Balarka

@MikeM: I've not seen Cliff Taubes in decades, but I always thought he was nice.

Balarka Sen

I explained Karim what's happening with the delta complex structure of the torus.

Ted Shifrin

So did I, Balarka :D I wanted him to do extra examples. We struggled on a sphere with 2 2-cells. Now he's supposed to be doing the torus when we cut it horizontally.

Balarka Sen

Aha.

Mike Miller

17:45

Thanks @Ted.

Ted Shifrin

I don't think he can manage to picture a (1,1)-curve on the torus to cut it in half along it ...

Balarka Sen

I drew the diagonal curve for him when he said he couldn't.

Ted Shifrin

@MikeM: General rule of thumb would be to bug reasonable experts closer to you first ...

You drew it for him on the torus? @Balarka

Mike Miller

@Ted: I don't know who those are.

Balarka Sen

Yeah.

Ted Shifrin

17:47

Cool ... @Balarka: I described it to him, wanting him to try to draw it, but that's fine with me.

Mike Miller

I guess Peter would be first.

Ted Shifrin

I doubt that, @MikeM. Peter does (did) totally different stuff.

Balarka Sen

I didn't know you assigned it to him, otherwise I wouldn't have revealed.

Mike Miller

@Ted: Then I think we're out of options. This is not Ciprian's style.

Ted Shifrin

I didn't assign it, Balarka. He put up the picture from Hatcher. I assigned him easier problems.

OK, @MikeM ... unless you know some younger guys who work on gauge theory stuff.

Mike Miller

17:49

@Ted: Matt (you met him) and I are thinking about it, but ASD metrics are somehow very different than ASD connections, and nowadays much less popular.

Balarka Sen

@TedShifrin As a side note, I haven't forgotten much of the calculus I have learnt.

Or so I think.

Ted Shifrin

@Balarka: What's the easiest example of a finite-dimensional CW structure that is not a $\Delta$ complex?

Balarka Sen

I went though a couple exercises from the book, and apparently I can do them.

I'll try the harder ones after I finish revising.

@TedShifrin S^2.

Semiclassical

@pval interestingly, from what i'm seeing numerically, the spectra of those matrices are approximately symmetric in distribution about $-1$

Balarka Sen

1 2-cell.

1 0-cell.

Ted Shifrin

17:51

Oh, I've forgotten the rules. You have to have codimension 1 faces along which to glue for $\Delta$?

Balarka Sen

Yes, you need to glue the faces of one simplex to faces of other simplex, preserving orientation.

Ted Shifrin

That didn't quite answer my question.

You're required to glue in codimension 1 ?

Mike Miller

@Ted: Yes.

This is more tautological than I think you'd like. Gluings in CW complexes are along maps; gluings in delta ocmplexes are along linear homeomorphisms.

Balarka Sen

@TedShifrin I don't know what a face of codimension 2 means. But yes.

Ted Shifrin

OK, thanks. I never learned about $\Delta$ complexes when I took algebraic topology, and when I've worked with the grad students on the topology qual we've just talked about cellular homology.

You have "faces" of all dimensions up to the top dimension, @Balarka.

Semiclassical

17:54

actually, i'm wrong (though not so wrong)

Mike Miller

That's the main difference. It's straightforward to see that you can subdivide a CW complex into a Delta complex iff the CW gluing maps are embeddings

Ted Shifrin

@MikeM: So attaching maps can only have degree $\pm 1$?

Balarka Sen

@TedShifrin Oh, I don't call those faces. But I see what you mean.

@TedShifrin Yes.

Ted Shifrin

What do you call 'em, @Balarka?

Well, now we all know why I've never been a topologist :D

Mike Miller

@Ted: You're thinking of them wrong. We don't attach by the boundary. We attach by the faces.

Ted Shifrin

17:55

@MikeM: I much prefer cellular :D

Balarka Sen

@TedShifrin Nothing particular. Simplices. :)

Ted Shifrin

But, still, I can triangulate a ball and think about building $\Bbb RP^n$.

Balarka Sen

$\Delta$-complexes are a semi-linear version of CW-complexes. The most linear version is simplicial complexes, where each simplex is uniquely determined by the faces.

Ted Shifrin

Damn, Facebook sure gets annoying on birthdays :P

Balarka Sen

Or at least, that's my personal philosophy.

@TedShifrin Oh, happy birthday!

Mike Miller

17:57

@TedShifrin People have a hard time with the pictures.

Ted Shifrin

Thanks. Well, triangulations are a pain. In diff geo when we do Gauss-Bonnet, I make my students triangulate a torus and a 2-holed torus. It takes some ridiculous number of faces ...

Mike Miller

Yeah.

Balarka Sen

@TedShifrin The local degree formula is a nontrivial bit though. I use to think that's why Hatcher introduces cellular homology late.

Oh yeah triangulations are shit.

Ted Shifrin

Yes, @Balarka. It's again why I would prefer to teach alg top to people who already knew some diff top. :)

Balarka Sen

I think the minimal triangulation on a torus requires like 27 faces.

Ted Shifrin

17:59

I think it's 18, @Balarka.

Mike Miller

@Ted: 14.

Ted Shifrin

Hmm ...

Mike Miller

Heawood graph

In the mathematical field of graph theory, the Heawood graph is an undirected graph with 14 vertices and 21 edges, named after Percy John Heawood. == Combinatorial propertiesEdit == The graph is cubic, and all cycles in the graph have six or more edges. Every smaller cubic graph has shorter cycles, so this graph is the 6-cage, the smallest cubic graph of girth 6. It is a distance-transitive graph (see the Foster census) and therefore distance regular. There are 24 perfect matchings in the Heawood graph; for each matching, the set of edges not in the matching forms a Hamiltonian cycle. For instance...

Ted Shifrin

OK ... Well, the obvious one for an undergraduate to find from the rectangle picture is 18 :P

Mike Miller

I think I once saw a proof that the triangulation I'm referring to is minimal.

Balarka Sen

18:01

I'd like to see that proof.

Ted Shifrin

Heya @AndrewT

Andrew Thompson

Hellu!

Balarka Sen

@TedShifrin Good thing about $\Delta$ complexes is that there is always a canonical homeomorphism with a simplicial complex, by barycentrically subdividing twice.

Ted Shifrin

Oh, I didn't know that. So if we take the simplest $\Delta$ complex for the torus and do this, how many faces do we end up with in the triangulation? 18, I'm guessing.

Andrew Thompson

Is it your birthday, @TedShifrin? If yes; happy birthday!

Mike Miller

18:03

@Ted: See the discussion on page 10 here.

JeSuis

@TedShifrin Bonsoir

Ted Shifrin

LOL, yes, @AndrewT. Thanks.

Balarka Sen

Eh. You have to $2$ 2-simplices. Barycentrically subdividing a triangle gives 6 triangles, right? So 1st step gives $12$. And the second step gives $72$...

JeSuis

@TedShifrin Joyeux anniversaire!

Mike Miller

Everything is determined by the number of vertices, and you can minimize the number of faces by minimizing the number of vertices. There's a bound: $V \geq \frac{1}{2}\left(7+\sqrt{49-24\chi(M)}\right).$

Akiva Weinberger

18:05

@TedShifrin Just got your reply. — Wait, I thought you said that each half has only half of a longitude, or something like that.

Ted Shifrin

Merci bien @JeSuis.

Akiva Weinberger

Am I confusing longitude and meridian?

Mike Miller

So it suffices to construct a triangulation of the torus with 7 vertices, which is possible.

This is the minimal picture.

Semiclassical

@pval what i saw specifically is that for large matrices, the eigenvalues appear to be symmetrically distributed around -1. however, there's a a fraction of them (about 1/n) which occur at negative values and which don't have a positive partner

Akiva Weinberger

Longitudes are the horizontal lines, right? The ones that vary in size? And the meridians are the vertical ones

@TedShifrin

Andrew Thompson

18:07

I can never keep track of what is horizontal and what is vertical.

Why can't people say 'up-to-down' and 'left-to-right'

Ted Shifrin

Backwards, @Akiva.

Akiva Weinberger

@TedShifrin math.oregonstate.edu/~garity/333S13/Notes/13_Section5.1.pdf

Mike Miller

@Ted: That's not backwards to me.

Ted Shifrin

Think about latitude and longitude on the sphere, guys.

Balarka Sen

I thought meridian were the vertical ones.

Akiva Weinberger

18:09

@AndrewThompson "Horizontal's down the hall, vertical is up the wall"

Mike Miller

@TedShifrin No.

Balarka Sen

i.e., the ones which bound disk.

Akiva Weinberger

On a sphere, meridians and longitudes are the same thing, I thought

Andrew Thompson

@AkivaWeinberger As both were described with 'up' and 'down' it doesn't make sense to me.

Balarka Sen

@AndrewThompson I can never keep track of what is up and what is left.

Ted Shifrin

18:10

I call them profile curves and parallels to remove ambiguity.

Akiva Weinberger

Um

Ted Shifrin

Meridians are parallels.

Semiclassical

taking the case of $2n=40$, for instance, Mathematica gives the following histograms (the first for the entire range, the second for just 'near' -1):

Ted Shifrin

@Akiva: Just because you find it on the internet doesn't make it correct. People post wrong crap all the time.

Akiva Weinberger

But they're both parallel, in the sense of not intersecting others of the same type.

Ted Shifrin

18:10

For a surface of revolution, parallel means literally parallel.

Balarka Sen

@TedShifrin Confusing, I thought parallels mean the longitudes because they run parallely to an imaginary center circle of the torus.

Conclusion: You cannot remove ambiguity.

Ted Shifrin

No, @Balarka. They are literally parallel in Euclidean space.

Balarka Sen

@TedShifrin Right, I figured after you said they were meridians :P

Akiva Weinberger

Agh

Semiclassical

Ted Shifrin

18:12

@MikeMiller Very cool. Thanks.

Balarka Sen

lol, we have successfully confused @Akiva.

Ted Shifrin

Now the dog really will eat his homework.

Mike Miller

@Ted: In topology, a longitude of a knot is a curve on a small torus around it that has zero linking with the knot. A meridian is a curve on said small torus that has linking 1.

Akiva Weinberger

Wait, doesn't that agree with my link?

Mike Miller

Yes.

Akiva Weinberger

18:14

I.E. Meridian=small ones

Ted Shifrin

So that appears to be backwards from the geometry convention, @MikeM. Fuck you topologists.

Balarka Sen

that is standard terminology, iirc.

lol.

Semiclassical

conventions, i shake my fist at thee!!!

Mike Miller

@TedShifrin Longitude, as every topologist knows, stands for "long".

Akiva Weinberger

Maybe we should call them longitudes and shortitudes

Ted Shifrin

18:15

Topologists don't even know what "long" and "short" mean.

Balarka Sen

But I can make the meridians long too.

Semiclassical

i'm more used to the terminology of 'transverse' and 'longitudinal'

Akiva Weinberger

But which is which?

Ted Shifrin

I'm sticking to profile curves and parallels.

Balarka Sen

This is fun.

Ted Shifrin

18:16

I'm tired of this.

Balarka Sen

Oh, heck, I forgot I had questions to ask, @Ted.

Akiva Weinberger

"Transverse" makes me think of a universe that's open-minded to people's sexualities

Ted Shifrin

Unless "verse" is short for "averse" :(

See, terminological problems abound.

Akiva Weinberger

Why does it even have the word "logical" in it

Balarka Sen

@TedShifrin In question 11, page 120, your function $F$ is not even defined on $v = 0$. Is that the reason of the paradox?

Akiva Weinberger

18:19

But is it defined on a meridian?

Ted Shifrin

Tell me which section, @Balarka.

Akiva Weinberger

OK, I'll stop.

Balarka Sen

@TedShifrin 3.6. Higher partials.

This is the first question on the list, there's more.

If $F$ is not defined on $v = 0$, you can't get it to be defined on a square around the origin, so the solution done before doesn't apply.

Ted Shifrin

I think you're wrong about the not defined on $v=0$.

Akiva Weinberger

You know, if we embed a torus in a 3-sphere, you can't even distinguish between the two. Its insides and outsides are homeomorphic (or even congruent if the torus has half the volume of the 3-sphere's surface volume).

Mike Miller

18:23

@AkivaWeinberger: One talks about this in the context of knots.

Or geometry.

Ted Shifrin

Correct, @Akiva. For the Clifford torus (which is what you're talking about), it's got zero curvature and is symmetric.

Mike Miller

In which case they're distinghuishable.

Akiva Weinberger

3-spheres are geometry

I said "congruent"

PVAL

@AkivaWeinberger Depends on the embedding I think.

Ted Shifrin

@PVAL: The standard Clifford torus :P

Balarka Sen

18:25

@TedShifrin Aha, right, I misread an or as and.

Ted Shifrin

LOL ... Only in English could you have a sentence saying an or as and.

Akiva Weinberger

I was just about to comment about that

Balarka Sen

@AkivaWeinberger I can even make the (1, 1) curve indistinguishable from the meridian, I think.

Lots of self-homeomorphisms of the torus.

Mike Miller

@Akiva: 3-spheres considered as what kind of object?

Akiva Weinberger

A subset of $\Bbb R^4$?

Ted Shifrin

18:26

Or as abstract Riemannian manifold of constant curvature :P

Akiva Weinberger

@TedShifrin Wrong

Spanish has "un o como y"

:P

Mike Miller

What do you consider the automorphism group of the 3-sphere when it's a subset of $\Bbb R^4$?

Ted Shifrin

Actually, I would have typed saying an "or" as "and." :P

Akiva Weinberger

@MikeMiller Its rotation group? Or something like that

Balarka Sen

@TedShifrin Wait, I am still not convinced. Is $F$ really defined at $(0, 0)$? I don't think so.

$u, v$ are neither $< 0$. Also, even if $u \geq 0$, $v > 0$ is not satisfied.

Ted Shifrin

18:30

No, it's not defined at the origin, @Balarka, but you still have a path-connected space.

Akiva Weinberger

@MikeMiller Also its reflections, I guess.

Mike Miller

Ok, so you're really considering it as a Riemannian manifold, or possibly as a metric space. I believe you now that it's geometry.

Balarka Sen

@TedShifrin But how you found the solutions to $\partial^2 F/\partial u \partial v = 0$ requires that $F$ is defined on some rectangle, because you're integrating. $F$ is not defined on a rectangle around $(0, 0)$.

Ted Shifrin

So what?

You can get from anywhere to anywhere with little rectangles.

Balarka Sen

Oh, good point. Hmm. But then the $\phi, \psi$ would vary rectangle to rectangle.

I need to think.

@TedShifrin Yeah, so then, $F$ would be defined piecewise, i.e., rectanglewise. On each rectangle, it'd look like $\psi(u) + \phi(u)$, but not globally so, right?

That's what's happening to your $F$.

Ted Shifrin

18:37

Why don't they patch?

Balarka Sen

@TedShifrin That's an interesting question. I feel like there's a specific reason but I can't come up with one.

This is a cool problem.

Ted Shifrin

It's analogous to the beginning calculus issue of $f'=0$ on the domain tells us $f$ is constant. Or does it?

Akiva Weinberger

May I ask what the problem is? Out of curiosity

Ted Shifrin

Make sure you don't do this one sideways, MyDogAte :P

Balarka Sen

@AkivaWeinberger I'll let you know what the problem is after I solve it :P

@TedShifrin It doesn't. $f$ can be $1$ on $[-1, 0)$ and $2$ on $(0, 1]$, right?

Akiva Weinberger

18:47

@TedShifrin Columbus isn't even the name of my dog. Her name is Matana.

"Columbus ate my homework" was just nonsense :P

Ted Shifrin

Right, @Balarka. LOL, well, I tried, MyDogAte :P

Balarka Sen

I must admit, your calculus book is great. It's relatively much more enjoyable than Hatcher (relative means enjoyability modulo the vastness of the subject).

Ted Shifrin

Hatcher is far, far, far ... more sophisticated.

Idle001

salut @TedShifrin

Ted Shifrin

salut @Idle :)

Balarka Sen

18:51

Hatcher can be more enjoyable, but algebraic topology is way more vast than calculus. That's why I said relative.

Idle001

vous vous portez bien monsieur @TedShifrin ?

Ted Shifrin

Oui, merci bien, M @Idle :)

Idle001

ravi de le savoir monsieur ted

Akiva Weinberger

I just tried looking up "calculus textbook by shifrin," but Amazon keeps giving me this text by some guy named Theodore? :P

Idle001

not theodore the bomber right ?

Ted Shifrin

18:54

LOL, après vous, Georges Dandin.

Balarka Sen

In case you haven't noticed, they're probably the same person, @Akiva.

Akiva Weinberger

I was joking.

Hence the ":P"

Balarka Sen

@Idle001 lol I had forgotten unabomber's initial name was Theodore too.

@TedShifrin Shall I ask the second question?

Idle001

i m not going to forget him and his wierd solution for society for sure

Ted Shifrin

Go ahead, Balarka.

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$Mathematics$ Mathematics