You were talking about the proof...

@JonasTeuwen $\sqrt{x^2+y^2}$ is not $\mathcal C^\infty$ at $(0,0)$, but I eventually realized that the proof doesn't need that

@FrankScience I know.

Holy monkey, I was just saying that you can redefine it to be.

@JonasTeuwen You meant that the proof is easy?

No, but it actually is, yes.

@JonasTeuwen Write your proof please.

Of the statement?

@JonasTeuwen Yeah, about the extension.

@FrankScience Oh, right. For each derivative you redefine the point in $(0, 0)$ to be the limit (which will exist).

And then you show this gives a type of function you need.

What's $(0,0)$?

The moral is that if it only goes wrong in one point, fix it there!

Here the point where it goes wrong.

This statement

Yes, that is what it says.

It states what it means for a discontinuity to be removable.

I gave the example $\frac{x}{x}$.

No

@FrankScience Oh okay, then I shut up.

For example, $\mathcal M=\{(x,y): x^2+y^2<1\}$.

We should extend the function into $\mathbb R^2$

I'll post a question. If you think you're right, post there, it's better for the others.

Oh, slight different statement.

So basically what you should worry about is the boundary.

@JonasTeuwen : )

@FrankScience But if the function is not analytic, this extension will certainly not be unique.

Yes, of course.

Edge is probably "boundary"?

yes

@FrankScience I mean, you can do a pretty explicit construction I suppose. You know the function $e^{-1/t}$?

Only prove that the extension exists.

But then you must play a bit with it, but there are also some general theorems.

Yes, but give one and you proved it.

So you only need to have one in some small $\epsilon$-ball near your boundary.

Why only need?

Because then you just make it some constant or whatever.

So put something away from your boundary and now connect it smoothly.

Well, I'll post one question, and if you think you're right, please publish, because there might be many people who wonder how to prove it.

It is not that easy.

Because you give a general boundary.

I think it's difficult, at least the proof on the book seems so complicated.

Which book?

Григорий Михайлович Фихтенгольц

Ah, right, I'll check the book shelf immediately...

You're Russian?

No. Is joke.

There seems no English translation of that book.

Looks like some Spivaky-thingie.

I mean, why would you bother doing this. Unless you like geometry.

But I do know, that in general it is quite hard.

But all the cases I have seen are easier because you have for example say that circle.

The book pointed out that you'll see the importance of these theorems in succeeding chapters.

What is this book about?

Calculus.

Notice that $\mathcal M$ might be disconnected, but it's unimportant.

Aha. Calculus. Oh my, those Russians.

Oh, my what?

Well, usually you could do stuff like defining: $\psi_\epsilon(x) = \epsilon^{-n} \exp \left (\epsilon^2 (\epsilon^2 - |x|^2) \right )$. if $|x| < \epsilon$ and $0$ elsewhere.

@FrankScience Quite hard for calculus.

Then you can take the convolution with some nice indicator function which is nicely one near the boundary.

Then you get like a smooth cut-off function.

Incidentally, which one is more appropriate: $f$ is in $\mathcal C^n$; $f$ is $\mathcal C^n$.

I want to post a question now.

Hmm... I don't know. Both sound okay, right?

Oh, I forgot the $^{-1}$ in the second term in the exponential.

I see an article in mathworld is that, $f$ is $\mathcal C^\infty$.

Maybe one needs to write $f$ is in $C^\infty(\mathbf R^d)$.

Here now

@FrankScience Oh! I think you do my argument, basically. What do they do in the book?

The interesting thing is that there is a neighborhood of every boundary point, hence you can be a bit outside of it.

Some strange constructions.

@FrankScience But, the hypothesis basically says, for any point on the boundary, we can extend it a bit further, right?

yes

So, what do you do? You say! Cool, this is excellent. Take the extension for each point, then take some stuff in between the "end" of that new boundary.

And then function domain itself.

So, that will be outside the boundary, but still in a smooth part.

Post your answer, and as intuitive as possible, and as rigorous as possible, and as elementary as possible.

So, you take a cut-off function which is 1 up to that point.

I am explaining it to you now!

Good.

@FrankScience Wrote down some intuition.

Adding more.

Haha, extension by the axiom of choice.

Hello bro.

Sup.

How are you?

Me? Okay. You?

Ok. Except for: I should be studying (hard core) but haven't done anything so far today because sort of bonged out by imposed stress.

shrug

What course?

Either Futile Attempts (haven't started with that) or more Atiyah-MacDonald (CA).

Excellent.

Doesn't really matter which I do today, just need to do something.

Ready? Set... go!

x_x

In a minute. I think I need moar cofffee first.

@JonasTeuwen I miss the condition: $n\ge1$.

@JonasTeuwen Did you see the bronie pictures I upoaded on fb? I got >20 likes. 0_o

Someone wrote "Haha, ponies better than cats" : )

@MattN I did not.

My legs still feel funny.

@Jonas @Frank : In the definition, for the meaning of the word 'extension' to be fully justified, don't you think domain of $f$ should be a subset of the domain of $f^*$?

@MattN :P.

@RajeshD It's not necessary.

then why do you call it an extension?

@RajeshD For ease.

lol

@RajeshD Otherwise, I should explain more about the extension about neighborhood.

@JonasTeuwen We had this running gag "WE CANNOT STOP THE HEEERD!!" because one of the riders (has her own horse and does 3 day eventing) fell off on the first day. I didn't see it because I was at the front of the herd (we had 50 horses with us all the time and need to herd them) but apparently she first lay still for 1 minute because that's what you're supposed to do (at least in the US apparently, to check whether you're ok) and meanwhile the herd and her horse kept on running.

So one of the guides shouted: We cannot stop the heard! (instead of "Hey, are you alright?")

And she had to run after her horse and the herd : D

(the terrain is steep and stony and she must've been exhausted after all the running)

@RajeshD there's a neighborhood into which $f$ could be $\mathcal C^n$-extended. If I use your definition, the decription should be horrible.

It was very funny.

@MattN But what about the PONIES??

@Frank : I am not saying to use my definition, I just found the use of the word 'extension' a bit odd

@JonasTeuwen You're supposed to call them horses. : )

@MattN They will always be ponies for me.

Even though they are cute and small.

@RajeshD And I only used the definition on Григорий Михайлович Фихтенгольц

@JonasTeuwen Yes. But Icelandic people might be offended if you call their horse "pony" : )

@RajeshD That's not my own definition.

I got

@MattN Complain they should not, they are very tiny themselves.

@JonasTeuwen : )

@JonasTeuwen Subcover is necessary, by Borel's lemma.

@FrankScience Huh... this was calculus right?

If you allow me to use a hammer it is trivial.

Oh dear. I should go and do some studying.

See you later.

@JonasTeuwen Yes, calculus.

@JonasTeuwen Borel's lemma about open set cover for the closed set.

If this is "calculus" I wonder what the course on PDEs will be about.

@JonasTeuwen Yesterday I've used that lemma.

@FrankScience I just typed the idea I would have how to solve it, not a solution.

@JonasTeuwen That book's title, in English, might be: The tutorial for calculus.

@JonasTeuwen see here

@MattN Good luck, pony.

@JonasTeuwen Borel's lemma is only a side effect of the completeness of $\Bbb R$.

That's a different one than the one I had in mind.

Hi @jonas. I just saw you and tb were discussing about a problem, could you post a link of that, or tell me in words (in summary) of what it is about?

If you have seen it, you have seen it all.

@JonasTeuwen You have apply the compactness of $\Bbb R^2$ on the $\mathcal L$?

@FrankScience Compactness of ... what?

I want to have a finite subcover of the boundary.

Scientists say that the world is made up by protons, neutrons and electrons. They forgot to mention morons.

However, what you often can do is some compactness argument and take a finite subcover and you will be great.

@JonasTeuwen From your answer

@FrankScience Yes, of the boundary.

The closure of your domain is compact.

@JonasTeuwen the calculus book not only on boundary but also for inner points (inner point's neighborhood is obviously extending $f$). Such extending seems a joke.

@FrankScience Which extension seems a joke?

Hmm. "How do you select a Borel representative on $L^2$?" this stuff is out of my range, I give up on this

Extending the function to a point inside its domain?

@JonasTeuwen Extend a function into its subset of domain.

@FrankScience Cool.

Maybe they mean inside a subset that includes the boundary.

@JonasTeuwen In my definition, it's okay.

@RajeshD $\Bbb R^2$ is compact, so if the set of open sets $\Sigma$ covers a closed set, a finite subset $\Sigma^*$ can do, too

@Frank : check this

that was not in reply to your previous message

@FrankScience I don't have much idea on these things

@RajeshD I got it before. But the definition from calculus book seems prettier.

@JonasTeuwen I'll try to read the calculus book again. More efforts are necessary.

@FrankScience Why is $\mathbf R^2$ compact? It is only locally compact to my knowledge.

@JonasTeuwen With the trivial topology is is :D

@BenjaLim Then like everything is, omg.

Never knew this omg, yesss! Thanks. Great insight.

@JonasTeuwen Didn't know $\Bbb{R}^2$ with the trivial topology was compact????

I will show you something really stupid

Of course I did, Sherlock.

Put the trivial topology on $\Bbb{R}$

Please do! :-).

I'm all eyes.

the sets $(0,\frac{1}{n})$ are compact

@BenjaLim Only calculus is needed.

@BenjaLim really?

their intersection is empty :D

Aha... Excellent.

@JonasTeuwen Do you know Bolzano's method?

@robjohn The only open cover of $\Bbb{R}^2$ is.... $\{\Bbb{R}^2\}$

@FrankScience The what?

@BenjaLim We are all trolling you.

@JonasTeuwen For example, cut $[a..b]$ into $[a..m]$ and $[m..b]$, and so on.

@JonasTeuwen repeat cutting,

@JonasTeuwen the example above does not stuff up when you space is at least $T_2$

@BenjaLim I was thinking of the topology where each point is an open set.

@JonasTeuwen $m=(a+b)/2$.

Oh? Cutting things in pieces? I work on harmonic analysis, I cut up things real good all the time! 8-).

@robjohn discrete topology?

@BenjaLim Oh yeah, filterrrrs.

@BenjaLim yeah, I was thinking of the wrong one.

@JonasTeuwen OK, you can try to cut a rectangle into four picses, and so on

@JonasTeuwen What about filters?

@robjohn Hey ya know what

It feels so cool to just write down $\pi_1(X,x_0)$ :D :D :D

They have the finite intersection property too.

@JonasTeuwen I don't know what is a filter...

It is a net!

In some way.

I need to read about that

@JonasTeuwen Do you know much about CW- complexes?

And that... is like a generalized sequence.

Not much.

I only work with cubes man.

And balls.

Cutting up open sets.

ahahahahahahah

in hatcher

CW - complexes

I just got mind - ******

@JonasTeuwen Supposing there's one counterexample, say, $\mathcal M$ necessarily need infinity many open sets to cover, and $\mathcal M$ is in a rectangle. Try to apply Bolzano's method on the rectangle.

@robjohn wanna know some personal philosophies?

How is physics forums? Is it good?

@t.b. OMG WHERE HAVE YOU BEEN

@FrankScience Huh? No, it says that every open cover has a finite subcover, right?

@t.b. NEXT SEMESTER ALGEBRAIC TOPOLOGY

So cover it with balls! 8-).

@JonasTeuwen Yes. We should prove it, so we suppose it's wrong, and discreminate.

Yes, so I give you one open cover where you cannot select a finite subcover.

@JonasTeuwen $\mathcal M$ is the closed set where infinite cover is necessary.

@JonasTeuwen $\mathcal M$ is bounded by a rectangle, and $\mathcal M$ is cut, as the rectangle is cut into 4 pieces.

What the monkey are you talking about?

An open set such that the boundary cover does not have a finite subcover?

@JonasTeuwen What are you guys talking about?

@JonasTeuwen A closed set doesn't have a finite subcover.

@BenjaLim I do not know.

@FrankScience If it is bounded it does.

@FrankScience Sounds like some compactness stuff I can help out with

I don't know what he is doing: 1) Trolling me 2) Disagreeing with me I actually agree with 3) I don't know.

@JonasTeuwen Here is something you may be interested to know

@JonasTeuwen You said that you cannot prove it in calculus, or I misunderstood your meaning?

@FrankScience Yes, you misunderstood.

I find daily life so mundane and boring compared after looking at the depth of maths :(

@JonasTeuwen Seems like stuff is so mundane; people go shopping, smoke pot, drink, etc

@BenjaLim And you don't even know what a filter is yet. So it will become worse!

but smoking pot man after a while it gets boring

@JonasTeuwen Sorry.

i.sstatic.net/Y7Su4.png ..

@FrankScience What are you talking about?

@Manishearth Excellent because it is excellent!

@Manishearth edit yours.

@FrankScience Edit my what?

Oh right, forgot

@BenjaLim I was just proving a finite subcover exists, in elementary calculus, sorry.

@FrankScience For....?

@FrankScience Are you to trying to prove compactness of $[0,1]$?

@BenjaLim I misunderstood.

@BenjaLim It doesn't make sense.

@FrankScience Compactness of $[0,1]$ does make sense

@BenjaLim I meant that the work I did doesn't make sense.

@FrankScience what were you trying to do?

I have no idea about the topology.

@FrankScience Doesn't matter, could you tell me what were you trying to do before?

@BenjaLim Try to explain the proof which I know to prove that the finite subcover exists.

@Manish : Where did you get the image from? have you made it up

@RajeshD Nope, one sec

gravatar.com/avatar/…

@JonasTeuwen You are working on harmonic analysis?

Found a link to that on a comment on MSO, can't remember where

ok

good one

@JonasTeuwen I wonder whether I can understand the Poisson's summation formula when I finish studying the calculus?

Sure, it is very easy?

I don't know.

I saw it on CMath.

about the asymptotics for $\sum_k\exp(-k^2/n)$.

What is CMath

@JonasTeuwen It's proved by Poisson's summation formula, which is related to Fourier series that you must be familiar with.

@RajeshD CMath? Ok, Concrete Mathematics.

@RajeshD Covered with Elementary Mathematics.

ok

its a book

@FrankScience I am familiar with those, yes 8-).

@JonasTeuwen I first heard that it's related to asymptotics, so I get interested.

I never seen it being used that way. Reference?

CMath EXERCISE 9.59

@BenjaLim sure (sorry for the delay, I was answering a question and got into a comment spree :-)

@JonasTeuwen Somebody has posted an answer, here

Did anyone watch spider-man?

@FrankScience He does... the same 8-).

@JonasTeuwen Only an implement?

@FrankScience Yes.

@JonasTeuwen But I don't know how $\psi$ and $\phi$ constructed?

He explains this right?

@JonasTeuwen For technical reasons

@JonasTeuwen To say roughly, keep $f$ in $\mathcal M$, and make the outside area of $\mathcal L$ smooth enough.

@FrankScience Yes.

@JonasTeuwen I can't get some details, for example, he claims $\phi_k$ exists without proof. It suffices to make $\phi_k$ is $\mathcal C^n$-function, which I know how to construct, but what about his stronger result.

Huh, he just defines it.

Why is it essential that $\sum_k\psi_k=1$.

Well, to prove $f_*$ is smooth.

@JonasTeuwen His proof is prettier than the one on calculus book.

@Jonas Is there any restriction/condition on Fourier series coefficients of a periodic (with period $T$) function $f$ which vanishes on an open interval, (it can be closed too) $(a,b) \subset (0,T)$?

They are all $0$.

yes

$f$ vanishes only in a open interval which is a subset of $(0,T)$ not on entire $(0,T)$

@Jonas

Yes.

hmm

how come?

Hmm, $f$ is the Fourier series?

$f$ is the function

Let $a_k$ be the Fourier coefficients

Are subgroups of finitely-generated finitely-presented groups necessarily finitely-generated?

render

I cannot still understand the details.

Is it true that:

If $f$ is smooth on $[a..b]$, and smooth on $[b..c]$, then smooth on $[a..c]$?

No.

I need to go.

So how to check whether it is smooth for a function which breaks into many cases.

@FrankScience : From my experience, when Christian blatter answers it takes considerable time/ effort to understand it, he generally makes things compact. Reread it carefully and patiently

@RajeshD smooth is really out of my ability.

nothing

@RajeshD it's not introduced a lot in calculus.

a function is differentiable infinitely many times then it is called smooth

Yes, but I don't know its property.

feel free to use wiki pages to know about concepts which you have not yet encountered

that is its property

I know it

but I don't know the details

??

there is no more details

thats how things are

I remember he answered a question of mine.

Do you know bump functions?

no

@RajeshD Not necessarily. It just has to be differentiable a "sufficient" number of times. So for instance, $C^2$ splines would be considered "smooth". But I suppose it depends on the context.

@JM Yes

yes @JM I did encounter people using sufficiently smooth by smooth

@JM his answer

@BenjaLim Don't knock the pot. I'm, uhurm, told that it helps with visualizing $n$ dimensions... ;)

Calculus seems a bit complicated.

@FrankScience That's what they said too, when Newton and Leibniz first came up with it. ;)

@JM They are none of rigor.

@JM Hi there! You're becoming a regular ;-)

@FrankScience That question is more analysis than calculus, I think. ;)

@JM It's from a calculus book.

@robjohn There should be an "again" in that sentence somewhere. ;P (Also, hi!)

@JM I agree

Anyway, I should understand it.

@JM I was going to add "again", but I opted for the wink instead.

@FrankScience ...and we see that sometimes books are horribly mislabeled. :)

@JM Oh, got it, but I realize that it's not easy to understand.

does a real analysis tag need to be added to this question @JM?

@RajeshD That's my opinion, at least...

"bounded open set", Do you think these are introduced in calculus books?

@RajeshD I guess it depends on what "calculus" books are we looking at... :D

@RajeshD Yes, in my calculus book, when describing multivariables.

ok

I think a real analysis tag has to be added to this, atleast I'd be happy to see it

Ok, I'll add

it also gives more exposure to attract some answerers

ADDed

Is real-analysis more than calculus?

I too always had that doubt, but I have given up on that after coming to math.SE, @Frank : BTW I am not a mathematician, its just my hobby, so don't always take me seriously

@FrankScience There have been lengthy discussion in this chatroom about this. It seems that people use real analysis in many different meanings. See, for example, here.

Ahoy =9

@MartinSleziak Oh

$g(x)=(x-1)(x-2)^2(x-3)^3$

8=1+7

So $g(x)$ has three distinct roots, where $x=1$ is a root with multiplicity 1, $x=2$ is a root with multiplicity $2$ and $x=3$ is a root with multiplicity $3$?

I am just a tad unsure what the correct way of adressing roots on the form $(x-a)^c$ is.

yes that's correct

Now I just need to find of saying that in Norwegian. I guess you can not help me there.