@Vrouvrou: Je t'ai répondu hier avec une réponse, mais tu ne m'as rien dit.

Et oui, il faut que $K$ soit convexe.

Hi demonical @Alessandro

hi Semiclassic

Hi @Ted

Hi @TedShifrin

@Semiclassical Let me check it out

Oh, hi @Balarka

i need some help with this math.stackexchange.com/questions/2550723/…

@Semiclassical you there ?

@Tuki: That differential equation looks suspicious. Shouldn't you be using the heat equation?

@Ted Shifrin I am not aware of what heat equation is. This is something what i've heard of but never looked up what it is.

You need partial derivatives with respect to $r$ (in this case) and $t$.

what he's got is Laplace's equation in cylindrical coordinates i.e. steady state

Aha ... The problem doesn't state steady state.

well, it doesn't mention time

and you'd have to mention the initial condition if you were going to have the time-dependent heat evolution

shrug

OK, fair enough. I leave it to you.

anyhow

@tuki if you've got a specific question, ask it

but I already gave you a fairly precise suggestion as to how to proceed. did you manage to get something out of it?

@Semiclassical I get in situation where i divide my zero when i try to solve for integration constants $c_1$ and $c_2$

there most be somewhere that i made a mistake but i cant find it

Okay. Show your work.

it's in "Edit" section at the bottom

you're making your life waaay too difficult

Question

what's $e^{-\ln R+C_1}$? @Tuki

@TedShifrin You might dig the geometry in the article I linked above for Balarka

$X$ is a connected metric space, I must show that it is not countable. I have no idea how to proceed.

"Cosines and Cayley, Triangles and Tetrahedra"

What theorems do you know, @orbit?

@TedShifrin That the rationals are countable.

Thanks, Semiclassic. I clicked on it and will look later. It looks a lot like what you were doing.

You know more than that, @orbit. What course is this?

@Semiclassical that would be $\frac{e^{c_1}}{R}$

@TedShifrin Real analysis 1. Chapters 1-5 in Rudin

right.

so what's $\displaystyle \int \frac{e^{C_1}}{R}\,dR$?

Which chapter are you in?

Heya, DogAteMy!

DogAteMy, I already showed Vrouvrou yesterday how to do that ... but apparently he didn't bother to read it. Grr.

@Semiclassical That would be $e^{c_1}\ln{R}$ ?

Well, these are exam questions that I'm doing to get ready. So we're done all 5 chapters. The section on countability in Rudin is a bit sparse...

almost. it's another indefinite integral, after all...

also wit the constant

c_2

@TedShifrin je n'ai pas vu votre réponse je suis désolée

right. so $u(R)=e^{c_1}\ln R+c_2$.

No, this is coming from continuity, @orbit.

merci de m'avoir aidé

Can you apply boundary conditions to that?

Ah, oui, j'ai écrit la réponse hier.

oh, I see the other error

look at your boundary conditions again.

C'est à vrai dire le théorème de Pythagore.

@TedShifrin Oh. Interesting...Lemme take a look

@Semiclassical should be at R=1 and R=2

right. can you restate the boundary conditions just so we're on the same page?

Typing more ...

@Vrouvrou What theorem is this trying to prove?

We know that $$ u(1)=100 $$ and $$ u(2)=0 $$. Also function $u(R)$ is only defined in range $1 \le R \le 2$

@Semiclassical This should be in the "Problem" section where the problem is defined

Oh, given a point $x$ and a set $K$, there is at most one closest point in $K$ to $x$?

@TedShifrin The image of a continuous mapping of a connected subset of a metric space is connected. I don't see how that helps.

Do you have an obvious continuous real-valued function to consider on a metric space?

DogAteMy: convex set $K$, yeah.

@TedShifrin Ah, yes the metric itself.

Where are $m$'s @Abcd?

@Tuki Okay. So what does that tell you about the solution we just found?

Yes, but that's on $X\times X$, orbit, and you probably want only $X$.

@AkivaWeinberger If $K$ is a convex compact set on a Hilbert space then there exists a retraction $\pi_{k}:E\to K$ which is $1-$ Lipschitz

@TedShifrin Is it ok, if I share my handwritten work? I am facing real trouble with too many sines and \dfracs :/

Right, convex @TedShifrin

LOL ...

@Ted Well, I could curry it. Fix a point in X, and consider the distances to that point.

There you go, @orbit.

@Abcd: So we have $\dfrac{\sin(\frac{m+1}2x)}{\sin\frac x2} + \sin ((m+1)x)$?

@Semiclassical I could solve with system of two equations ?

with $R=1$ and $R=2$

sure. that's a bit overkill, though: what's u(1) according to your solution?

@TedShifrin Yes, and I simplified it.

@AkivaWeinberger theuy say as for all $x\in E$ there exists a unique $a_k\in K$ such that $||x-a_k||=\inf_{k\in K}||x-k||$ then $\pi_{k}(x)=a_k$ define a map which is the identity on $K$ , can you tel me why it is the identity ?

main thing is that there's no reason to ever be doing ln(0) = -infinity here

@Semiclassical i agree

you only evaluate ln(R) at R between 1 and 2.

@Vrouvrou If $x\in K$, then the closest point in $K$ to $x$ is $x$

$\sin \dfrac{(\dfrac{m+1}{2}x)\dfrac{\sin \dfrac{mx}{2}+ 2\cos \dfrac({m+1}{2}x)\sin \dfrac{x}{2}}{\sin \dfrac{x}{2}}}$

Hi!

$x$ is zero distance away from itself and positive distance from anything else

ah that's all

@Semiclassical and i need to know values for $c_1$ and $c_2$ right ?

Mathjax is terror !

i thinked that they prouve that $\forall x\in E, \phi_{k}(x)=x$

you'll be able to work them out, yes

So we're reduced to showing that $$\sin(\frac x2) \sin ((m+1)x) = \sin(\frac{m+2}2x)-\sin(\frac{m+1}2x),$$ and doesn't that follow easily?

@AkivaWeinberger

Terrible, or terror? @Abcd

Actually not only that proves that $X$ isn't countable, but also that $X$ has at least the same cardinality as $\Bbb R$, right?

in particular, you should be able to read off c2 from the condition at R=1

@AkivaWeinberger Both.

and that makes deducing c1 from the condition at R=2 pretty easy

Hi @AlexanderGruber!

@Semiclassical yes $c_2=100$

Sounds like @Alessandro took over with @orbit?

Afternoon friends

@TedShifrin i got something else, let me try to type it again :\

heya @Alexander

Long time. How's things?

Right. And c1?

@TedShifrin: Fix $x_0 \in X$. Consider $d(x_0,p)$, for $p \in X$. Since $X$ is connected and $d$ is continuous, we have that $im(d) \subset \mathbb{R}$ is also connected. Thus, we know $im(d)$ is some interval of $\mathbb{R}$. Since intervals of $\mathbb{R}$ are uncountable, and we have a bijection between $X$ and the interval, $X$ is also uncountable.

@Abcd: $\sin A - \sin B = ?$

Yep, that's a follow up question on Orbit's one

@orbit: Assuming $X$ has more than one point, yes :P

@AlessandroCodenotti yep

@TedShifrin Could I ask you a question (still regarding my problem of yesterday), given that you seem to be only of the few that really understands this stuff?

$\sin (\dfrac{m+1}{2}x)\dfrac{\sin (\dfrac{mx}{2}+ 2\cos (\dfrac{m+1}{2}x)\sin \dfrac{x}{2}}{\sin\dfrac{x}{2}}$

Things are a little busy, @nbro, but OK.

Ok, if you're unable answer, I understand

So, just to recall, the exercise I had was.

I remember it.

Anyway.

@TedShifrin oh shoot. If $X$ has only one point, then it is connected and countable...?

@TedShifrin, Finally, that's ^ what I reached.

This is what I've done so far.

@Abcd: You want to use the formula for $\sin A - \sin B$.

My problem now is that I am not sure how to derive $d_n$.

@nbro: I don't want to look at all that. I would recommend you first change the standard result from $[-\pi,\pi]$ to $[0,2\pi]$ (by extending the function periodically).

@TedShifrin is my simplification wrong?

@Semiclassical $c_1$ seems to be more difficult. I got $$ c_1=\ln(-\frac{100}{\ln(2)})=\text{non real solution} $$ it does have solution in complex plane though.

@BalarkaSen Decent. Writin' the ol dissertation

you continue to make things way more difficult than you need to

@Abcd: Surely you're missing a $+$.

your equation is $u(2)=e^{c_1}\ln 2+100=0$, yes?

@Semiclassical correct

so what kind of number does $e^{c_1}$ have to be?

@AlexanderGruber Cool

How bout yourself? Are you tenured and winning prizes yet?

@TedShifrin I am sure I am not. I used Double angle formula for sin.

thanks anyway

Haha, no. Trying to wrap up high school

@Semiclassical I am not sure if i know what your trying to say

Turns out every compact metric space is the image of some continuous function from the Cantor set.

I think that's the wrong approach, @Abcd, although it should get to the same place eventually, I suppose.

@Semiclassical positive real number ?

@BalarkaSen you are in high school?

well, let's check. ln(2)=0.693, I think?

so you'd need $e^{c_1}(0.693)+100=0$.

@Semiclassical correct calculator can confirm this

is that possible for positive real $e^{c_1}$?

@TedShifrin I have done exactly as the author has done:

@orbit-stabilizer Yeah that's a beautiful theorem. I think in fact the continuous function can be assumed to be a quotient map

Re high school. Only barely

Oh, the problem is you wrote the wrong formula at the very beginning of this discussion, @Abcd. Scroll up.

I didn't think it was right.

@BalarkaSen well, shit. That's crazy. The only thing I knew in high school was what a derivative and integral was - and only barely at that.

@BalarkaSen the funniest part of the paper I linked, btw

@Semiclassical i would say it doesn't

right. you'd need $e^{c_1}=-100/(\ln 2)<0$.

@orbit-stabilizer I started a little early. Knowing doesn't matter much so it's not a superb head-start

@Semiclassical i see

@TedShifrin Sorry.

but, is that a problem? Let's plug that into your solution and see what we get:

Hey @Daminark

Did you see the ting I wrote for your thing?

$u(R)=-\frac{100}{\ln 2}(\ln R)+100$

@Abcd: So if you have the author's solution, what is your question?

Does that satisfy the boundary conditions?

@BalarkaSen You'll be appearing for board exams in march?

Yes.

@BalarkaSen I'm at the University of Minnesota. Specifically, I"m at its Minneapolis campus.

The prof who wrote that? He's at the Duluth campus.

@BalarkaSen okay I'm happy now

@Semiclassical Whoa lol

yep

And then once we have all those maps and show that to

nice nearly-local connection

And then once we have all those maps and show things commute, we'd then be using the five lemma guy?

Maybe you should collaborate with him on a project

@TedShifrin I can't understand author's next step which is:

Hi Demonark

@Semiclassical yes it does

(and put me on the acknowledgements of the paper you'll write togather lmao)

Hey @Ted!

@Semiclassical I dont know how i didn't see that in the first place. That you could do it like this

okay. Now let's check its derivatives: $u'(R)=-\frac{100}{\ln 2}\frac{1}{R}$ and $u''(R)=\frac{100}{\ln 2}\frac{1}{R^2}$

As I said twice, @Abcd, he's using the formula for $\sin A - \sin B$ :)

@BalarkaSen CBSE, ISC or West Bengal Board?

does that satisfy the required ODE?

@Daminark Yep

That's the idea

@Abcd WBB

The five lemma is like, my best friend, we're going for coffee later. By which I mean I'll probably be reading about it over coffee

Even I like the five lemma (or used to).

All the best!

Five lemma is fantastic

Thanks @abcd

@TedShifrin Okay, thanks for your help.

Do you see that that does it?

Hmmm, did Demonark really say that all open sets in $\Bbb R^n$ are homeomorphic?

That was a mistake - Big Smoke

Oh, a mistake.

Well, no point making little mistakes :)

@Ted so I know you didn't do quite as much research since you preferred teaching, and you seem to have leaned to asking questions that were of a more classical or differential nature, but I am curious what methods you usually found yourself using from related areas

I did my share of research, Demonark, for the first 25 years or so.

I still don't know how to prove convex subsets of R^n are diffeomorphic to R^n

@TedShifrin No...

I think that's impossibly hard, Balarka. We had a reference somewhere.

I think 0celo7 wrote down a proof at some point

There's an MSE post from him

@Abcd: So isn't it right that $\sin A - \sin B = 2\cos(\frac{A+B}2)\sin(\frac{A-B}2)$?

Yes.

Wow, 0celo contributed something good? :)

And lol regarding that, what I had in mind was that open balls are all homeomorphic. At which point I accidentally confused that with open sets

@TedShifrin The record should reflect that that comment was made in jest

Go big or go home? @Kevin

I mean even in the case of n = 2 I need to invoke Riemann mapping theorem or something

@Abcd: So $A=\frac{m+2}2x$ and $B=\frac m2 x$ gives what?

@TedShifrin He's into super-detailed analysis. Let me find it

It's in some textbook somewhere, @Balarka, but I don't remember where.

@TedShifrin misread that initially, derp

It's a good thing I've been teaching trig this fall. We have our midterm on Sunday and then I can forget all the stuff I don't ever care about :P

Amusingly, I have cause to appeal to the Sin*Sin version of that identity right now

@TedShifrin Done the proof! Ty.

What's it moving to next?

Right. AoPS likes abusing those kinds of formulas (product to sum, too) too much for my taste.

Oh, OK, @Abcd.

tbh the best way to do product-to-sum is just to Euler it imo

Trigonometry is complicated

@TedShifrin You don't like trigonometry?

@Semiclassical It does satisfy with both boundary conditions.

yep

so that's your solution.

I don't like torturous formulas, no. I'd rather use complex exponentials, which I assume we'll do later.

@Semiclassical should satisfy also all values in between boundary conditions ?

Demonark, were you asking me?

not sure what you mean by that. you've shown that 1) it satisfies the ODE, and 2) it satisfies the boundary conditions.

We did polar coordinates. Then complex numbers, roots of unity, etc., and then lots of linear algebra (actually talking about geometric things with linear maps!).

OK, lunchtime for me. Bye.

so that's a solution the problem as asked. (you could ask whether it's the only such solution, but that's probably beyond what you're being asked)

@Semiclassical i mean we could take for example $1\frac{1}{2}$ and it would satisfy the condition $$ ru''+u'=0 $$

@TedShifrin bye! Thanks for the help!

See you Ted!

Does that satisfy the boundary conditions?

@Semiclassical yes it does

uh

@TedShifrin Ok, I didn't find the post by 0celo but here

$u(R) = 1\frac{1}{2}$ satisfies $u(2)=0$ and $u(1)=0$ ??

@Semiclassical $$ 1 \le R \le 2 $$

...be precise, man

that's $R=1\frac12$

@Semiclassical yes

It's kinda cool because it says small exotic R^4's embedded in standard R^4 must be badly nonconvex

then you can evaluate your solution there and get the value of $u$ there, of course

but you have nothing to check that against.

so it doesn't really tell you much

@Semiclassical yes this is where i was going

ah.

the point is that, once you've found a solution which satisfies the ODE and the boundary conditions, you're basically done. (in a formal course one would probably appeal to some theorem in order to conclude that the above solution is unique, but that's all.)

@Semiclassical how could i prove that it is satisfied when $1 \le R \le 2$. Basically on all values in between 1,2 and values 1,2