Mathematics - 2014-09-18 (page 5 of 5)

Balarka Sen

10:00 PM

@MikeMiller!

Ted Shifrin

Oh no, is @Mike here?

Mike Miller

what'd I say about empty pings, @BalarkaSen?

Balarka Sen

It's not empty

Anthony

I'm not working with a circle, the question is given the unit square with it's usual product topology, and using the equivalence relation that the interior points are equivalent to themselves, and (1,t)~(-1,t) and (t,1)~(t,-1) and (1,1)~(-1,-1), find an embedding into $\mathbb{R^3}$.

Balarka Sen

it contained a '!'

Mike Miller

10:01 PM

fair

Balarka Sen

@MikeMiller How're the preparation for quals going?

Mike Miller

ugh

Anthony

@MikeMiller hola.

Balarka Sen

@Mike has decided to reply to everything in a single word.

Ted Shifrin

But that space, once you perform the identifications, is homeomorphic to a torus, @Anthony. That's what they wanted you to understand.

Chris's sis

10:02 PM

I wanna conjecture something very nice ...

Mike Miller

Not quite, @TedShifrin, the last identification modifies it a teeny bit

Ted Shifrin

doesn't speak further to @Mike

Anthony

Well I understand that, but he wants a 1-1 continuous function.

robjohn

@Chris'ssis yeah... I didn't say it would be easy ;-)

Mike Miller

Should be $S^1 \times S^1 \times I$ with two points identified

Ted Shifrin

10:03 PM

Oh, good point, @Mike. I didn't read.

Chris's sis

@robjohn $$\frac{J_0(2)}{1^2}+\frac{J_0(4)}{2^2}+ \frac{J_0(6)}{3^2} +\frac{J_0(8)}{4^2}+ \frac{J_0(10)}{5^2}+ \cdots= \zeta(2)-\frac{3}{2}$$

Ted Shifrin

No, that's not right, @Mike.

Mike Miller

lol

unit square, not unit cube

how silly

so it's just the torus with two points identified

Balarka Sen

hey i was going to post that

Ted Shifrin

No, those points were already identified.

Mike Miller

10:04 PM

lol

whoops again

Ted Shifrin

That last equivalence is redundant, @Anthony.

robjohn

@Chris'ssis That doesn't match the value in the answer to your question... I should check the value

Alizter

@TedShifrin yes. i don't usually eat sweets but my sister gave me one.

Mike Miller

glad to be of no help :)

Ted Shifrin

I hope your teeth don't all rot, @Alizter :P

I enjoy it when you're of no help to me, @Mike.

It's better than catching me in errors. :D

Chris's sis

10:05 PM

@robjohn It's just a conjecture ... that's why I won't post it on main ... ;)

Anthony

@TedShifrin I thought that, but it's in the problem. I still have no idea how to find a function.

Sab ಠ_ಠ

guys I'll switch off the laptop it's too distracting for me to work :P

I'll be back later :)

c ya'll

Ted Shifrin

You need the parametrization of the torus as a surface of revolution, @Anthony. If you don't know one, see my notes.

bubye @Sab

Ice Boy

later

Ted Shifrin

I need dinner and then to pack for a trip ...

Glad to know you're still alive, @Mike.

Ice Boy

10:07 PM

Bye Professor

Ted Shifrin

bubye, @skull.

Balarka Sen

Byes.

Ted Shifrin

bubye @Balarka

It must be past your bedtime, again.

robjohn

@Chris'ssis Mathematica computes it to be 0.14493406684822639603 which is not near $\dfrac{\pi^2-4}{6}$

Anthony

@TedShifrin I did... You just mean spinning a circle around?

Balarka Sen

10:07 PM

@TedShifrin And, I have school tomorrow.

Yes!

well, today.

Indeed @Balarka.

see ya

But I don't have a circle....

Alizter

10:08 PM

@TedShifrin I brushed very well afterwards :)

Balarka Sen

@Alizter go to sleep.

Alizter

@BalarkaSen uha

Chris's sis

@robjohn Could you correct my answer above? It's $-3/2$

Balarka Sen

throws a table at @Alizter

Studentmath

I'm off for the night too. \o!

robjohn

10:09 PM

@Chris'ssis Now it looks better

Ted Shifrin

When you identify the endpoints of $[-1,1]$ to one another, you have a circle, @Anthony.

Night @Studentmath!!

Chris's sis

@robjohn :D

Balarka Sen

i think i'd be leaving too

robjohn

@Chris'ssis I figured that before you posted it :-)

Balarka Sen

bye world, bu-bye cruel world.

Ice Boy

10:10 PM

everybody is leaving :(

Balarka Sen

snif

robjohn

@IceBoy have you showered recently?

Ice Boy

@robjohn that must be it

the ether

Mike Miller

I thought you were off, @Ted?

Ted Shifrin

I am.

Anthony

10:12 PM

@TedShifrin I mean, I have a whole bunch of circles...

Oh you're going... Alright.

Thanks.

robjohn

@Chris'ssis Olivier just used the series I was about to try

It works and gives your value

Chris's sis

@robjohn Yeah, nice answer. This is the way to go.

Mike Miller

@Anthony Can you prove that mixed partials are equal for me?

Anthony

You mean Clairaut?

Mike Miller

Yeah.

Anthony

10:13 PM

Why do you ask?

I don't think I can.

robjohn

@Chris'ssis I don't have enough experience with Bessel function identities

Chris's sis

@robjohn Neither do I. That one is a good beginning. :-)

Anthony

@MikeMiller ?

Mike Miller

@Anthony I can't, and I wanted you to do it for me!

Anthony

Oh.

I have a short proof of it I've been meaning to read over.

Mike Miller

10:19 PM

Give it here, buddy.

Anthony

math.ucla.edu/~ralston/pub/EqualPartials.pdf

Mike Miller

bleh

Anthony

Is it bad?

Mike Miller

It's not to my taste.

Ice Boy

It reminds me of this

Anthony

10:29 PM

@MikeMiller I see.

Ice Boy

just kidding :-)

Anthony

@MikeMiller I still have no idea how to do this torus problem.

Mike Miller

@Anthony Get a square and glue opposite edges. It's a torus.

Anthony

I know.

How do I define a function.

That maps from that quotient space into $\mathbb{R^3}$?

Mike Miller

With cosines and sines, I'd bet :)

Anthony

10:33 PM

Ted was saying a surface of revolution, and he said I have a circle?

But I have more than a circle... don't I?

Mike Miller

Consider how to embed a circle into the plane. Now be a bit crafty and embed the torus into the plane.

err. into $\Bbb R^3$.

good luck embedding it into the plane lol

Chris's sis

@robjohn Hippalectryon's problem is good to be kept in mind. It's an exceptionally nice problem, but it's over my level. It looks incredibly hard without that initial assumption.

robjohn

@Chris'ssis over your level? You've been doing some pretty hard sums recently. I bet you could have gotten it after some work.

Anthony

Embedding the circle into the plane effectively just means parameterizing a circle, right?

Mike Miller

yes

Anthony

10:38 PM

of what significance is the equivalence between the end points of the line?

Mike Miller

huh

Anthony

Is that just saying not to count one point twice?

Mike Miller

take an interval and glue together the endpoints

you get a circle

if you don't glue them together you don't get a circle

Chris's sis

@robjohn It's also true I didn't catch a good time on it, and I didn't work on it too much, but I expected more to get an idea (that didn't come).

Anthony

@MikeMiller I know in the stupid way that you need to glue the ends

But like

if I parameterize the unit interval, why do I need the equivalence between the end points to call it a circle?

robjohn

10:40 PM

@Chris'ssis happens all the time to most people

Mike Miller

@Anthony I have to admit I don't understand what you're trying to say

what do you mean by parameterize the unit interval

Anthony

Probably nothing correct. If we're embedding the circle into the plane, you just glue together end points of a line. How do you do that, with a function?

Mike Miller

@Anthony I think I've been too fast and loose with terminology about 'gluing'. What you just said doesn't make sense to me.

Anthony

Oh dear. When you say embed the circle into the plane.

You mean glue together points of a line segment right?

Mike Miller

No, I literally mean a map $S^1 \rightarrow \Bbb R^2$ whose image is homeomorphic to $S^1$.

Anthony

10:43 PM

I see.

Mike Miller

$S^1$ can also be described as $[0,1]$ with the endpoints identified... but you still need a map

e.g., $t \mapsto (\cos(2\pi t), \sin(2\pi t))$

Anthony

And the reason the endpoints need to be identified (besides the obvious physical reason) is so you don't get two points going to the same point?

Mike Miller

I don't understand what you mean. The reason the endpoints need to be identified is because the unit interval is not the circle.

The circle is homeomorphic to $I /\{0 \sim 1\}$.

We need the endpoints identified so that we're actually describing a map from the circle...

Anthony

I think what I'm asking is why is the circle homemorphic to I/{0∼1}.

Mike Miller

Because there's a homeomorphism between them? :)

Anthony

10:47 PM

I mean I know it makes sense, but where would I go wrong if I tried defining homemorphism between the circle and $I$?

Mike Miller

I defined a map up above; verify that it's continuous with continuous inverse

@Anthony Try it and let's see

Anthony

Well the map you gave, it's just not bijective, eh?

Yeah, wait, is it pretty difficult in general to prove that two things are not homeomorphic?

Mike Miller

In general, yes. In this case, there are nice tricks.

So you're right; the map I gave you is not an injection. But what about, say, the map restricted to $[0,1)$? Why is that not a homeomorpism? It's a continuous bijection.

Anthony

I assume its inverse isn't continuous :P

Mike Miller

Sure. Convince me it's not.

Anthony

10:51 PM

But you're hard to convince!

Lemme think.

I mean it's just arccos/arcsin, right?

Mike Miller

Not on this one! All you need to show me is an open set that maps to a not-open set...

(i.e., an open set whose inverse - under the inverse mapping - isn't open)

Anthony

Uh.

Mike Miller

How about $[0,1/2)$?

Anthony

I'm thinking. My brain isn't working.

that's the arc from [1,0 -1,0), eh?

Oh crap that's not open.

I see.

Mike Miller

Aye.

Anthony

11:00 PM

Alright, I'mma do this torus problem.

Mike Miller

Anyway, $[0,1)$ and $S^1$ aren't homeomorphic because the first is not compact, while the second is.

Anthony

Then the rest of my homework.

Well, sure.

Topology is too strong

Mike Miller

$[0,1]$ and $S^1$ aren't homeomorphic because if they were, say by a homeomorphism $f: S^1 \rightarrow [0,1]$, consider $x_0 = f^{-1}(1/2)$. Then $f$ would restrict to a homeomorphism of $S^1 \setminus \{x_0\} \rightarrow [0,1/2) \cup (1/2,1]$. But the first is connected, and the second isn't!

Anthony

How do I define the open sets on $S^1$ besides the fact that it's homeomorphic to [0,1] (0~1)?

Mike Miller

@Anthony $S^1$ is normally defined as the subset of $\Bbb R^2$ of points of norm 1, given the subspace topology

Then the image of $[0,1/2)$ is not open in the subspace topology, because any neighborhood of $f(0)$ must contain points on the bottom half of the circle as well as the top

Anthony

11:05 PM

What does the subspace topology mean again?

Actually I guess I can google that.

It's open sets are the open sets of the bigger space, excluding elements not in the set.

Mike Miller

@Anthony The open sets of $Y \subset X$ are those of the form $Y \cap U$, where $U$ is open in $X$

This is the weakest topology possible to make the inclusion map $Y \hookrightarrow X$ continuous, @Anthony

Anthony

I see.

Thanks @MikeMiller.

Pedro Tamaroff

11:30 PM

@DanielFischer

Mike Miller

@PedroTamaroff Prove that mixed partials are equal. Fast.

Pedro Tamaroff

Use Fubini.

Take a rectangle, integrate over that, done.

Mike Miller

Keep going.

Pedro Tamaroff

That's it.

Mike Miller

Integrate what, bro? Come on.

Pedro Tamaroff

11:31 PM

What are your hypotheses?

Ted Shifrin

$C^2$

hi mr @Pedro, @Mike

Mike Miller

What Ted said.

Ted Shifrin

This is an exercise in my book, so I think I know how to do it :D

Pedro Tamaroff

@MikeMiller Well, suppose that at some point $D_{i,j}f(a)\neq D_{j,i}f(a)$.

Mike Miller

Keep on.

Pedro Tamaroff

11:33 PM

There is then a rectangle where the functions don't coincide.

Ted Shifrin

nods

Mike Miller

Sure.

Pedro Tamaroff

Integrating and using Fubini gives equality, but this contradicts the inequality.

Mike Miller

Integrating what, now?

Pedro Tamaroff

$D_{i,j}f(x)-D_{j,i}f(x)$

WLOG we're in $\Bbb R^2$, $i=1,j=2$.

Mike Miller

11:34 PM

Sure bro.

Pedro Tamaroff

So we're good?

@TedShifrin Help.

Ted Shifrin

I'm still waiting for you to say hello :D

Pedro Tamaroff

Hello!

Mike Miller

And why does that integrate to zero, @PedroTamaroff?

Ted Shifrin

Just do it, @Mike.

Mike Miller

11:35 PM

:(

Ted Shifrin

Use Fubini + FTC

Mike Miller

Surely I'm allowed to be a lazy son of a bitch sometimes.

Ted Shifrin

nope

Help, what @Pedro?

Pedro Tamaroff

Say $R=[a,b]\times [c,d]$. Then $$\int_a^b \int_c^d D_{2,1}f(x,y)dydx=\int_a^b D_1f(x,d)-D_1f(x,c) dx= f(b,d)-f(b,c)-f(a,d)+f(a,c)$$ @MikeMiller

The same for the toher.

Suppose $f: \overline{B(0,r) }\to\Bbb C$ is continuous, and holomorphic in the interior.

Ted Shifrin

or don't you mean $D_{1,2}$?

Pedro Tamaroff

11:37 PM

$D_{2,1}=D_2D_1$

Ted Shifrin

oh, it does?

Mike Miller

Thanks for letting me be a lazy piece of shit, @PedroTamaroff

Pedro Tamaroff

That's how I write them.

Ted Shifrin

That's one reason I hate that notation :P

I write $f_{xy}$ to mean $(f_x)_y$.

Pedro Tamaroff

I want to show that Cauchy's integral formula still holds for $f$ in $B(0,r)$, @TedShifrin.

Ted Shifrin

11:38 PM

Sure, @Pedro, what's your prob?

Pedro Tamaroff

I am not sure where to start. On the other hand, I could prove Liouville's theorem.

Ted Shifrin

Huh?

Pedro Tamaroff

Using Cauchy's formula, I proved Liouville's theorem.

I mean, on the bright side...

Ted Shifrin

So is the issue to prove that $\displaystyle\int_{|z|=r} \frac{f(z)}{z-w}\,dz = 2\pi i f(w)$ rather than for $|z|=r-\delta$?

Pedro Tamaroff

Right, so should I take a limit of some sorts?

Ted Shifrin

11:41 PM

Seems right to me.

Pedro Tamaroff

Let me see how I can justify it.

Ted Shifrin

@Pedro: I haven't seen @Kaj around here in ages ... although he did show up in my office yesterday.

Pedro Tamaroff

@TedShifrin I gave him a problem some time ago.

Ted Shifrin

Good :)

The justification you're thinking about shows up in complex analysis a fair amount, btw.

Pedro Tamaroff

@TedShifrin Curve length, I guess? Something of that sorts?

Ted Shifrin

11:48 PM

hmmm ... whaddya mean?

Pedro Tamaroff

I don't know.

Let me finish.

Ted Shifrin

LOL ... I'll go do some packing ...

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