Mathematics - 2017-02-27 (page 3 of 5)

3:00 AM

A funner example: is the set $\left\{\frac{1}{n}\;:\;n \in \Bbb N\right\}$ compact?

Now, can you show that, for every $x$, there is an open interval $(a,b)\ni x$ such that $f$ is bounded on $(a,b)$?

Sure — choose an epsilon, and make the interval $(x-\delta,x+\delta)$ so that $f$ can't vary more than $\epsilon$. It's surely bounded there.

Zach Hauk

oh ya :P

Akiva Weinberger

(Since it is always between $f(x)+\epsilon$ and $f(x)-\epsilon$)

@ZachHauk So now let our cover be the set of all intervals on which the function is bounded.

Clearly, this is a cover, since we just prove every point is in one of these intervals (using the fact that the function is continuous).

So use the fact that $[0,1]$ is compact!

This gives us a finite collection of open sets on which $f$ is bounded!

Do you see how to conclude? @ZachHauk

Daminark

But yeah so the way the direct proof that $[a,b]$ is compact goes is basically like this

So your open cover is $O$

Akiva Weinberger

This will also explain why it fails on intervals like $(0,1)$ (that aren't compact). An example unbounded function is $y=1/x$.

Daminark

3:04 AM

Let $A$ be the set of $x\in [a,b]$ such that $[a,x]$ is covered by finitely many sets in $O$

Akiva Weinberger

Yup

Daminark

It's clearly not empty because $a\in A$

Zach Hauk

sorry what

Fargle

@AkivaWeinberger Or $\tan \pi x$.

Zach Hauk

i was playing a gane

let me scroll up and read

Daminark

3:05 AM

And it's bounded above by $b$

So there's some supremum, call it $c$

Fargle

This just makes me want to read Munkres again. goes off searching

Akiva Weinberger

@Daminark This supremum is either in A or it isn't.

Both of which lead to it not being the supremum very quickly…

…unless the supremum is $b$.

Yes?

Daminark

We know that $c\in A$ because if we take an open set containing it, then you have $(c-\epsilon, c+\epsilon)$

Which is in that open set

But then take $c-\frac{\epsilon}{2}$

ozarka

Consider the quadratic function $\mathbf{\frac{1}{2}x^TGx+b^Tx}$ in four variables where $$\mathbf{G}=
\begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix}$$
and $\mathbf{b}=(-1,0,2,\sqrt{5})^T$. Apply the conjugate gradient method to this problem from $\mathbf{x_0=0}$ and show that it converges in two iterations. Verify that there are just two independent vectors in the sequence $\mathbf{g_0,Gg_0,G^2g_0,\dots}$.

In this case, what is $\mathbf{g_0}$?

Daminark

We have that $[a,c-\frac{\epsilon}{2}]$ is covered by finitely many sets in $O$ since it's less than $c$

But then take that finite collection and stack on that open set around $c$ we were just talking about

So $c\in A$

Now, assume that $c < b$

Akiva Weinberger

3:08 AM

But then also $c+\epsilon/2$!

Is finitely covered by the same cover!

Daminark

Then that $c+\frac{\epsilon}{2}$ screws us up

So $c=b$

Akiva Weinberger

Yay

Daminark

Thus, $[a,b]$ is compact

Zach Hauk

I was gonna say that because it's bounded at every point we conclude that it's bounded but I don't know why that argument wouldn't work for open sets

and hopefully I'm not making myself look stupid here

Akiva Weinberger

@ZachHauk It's bounded on each individual set of the cover, right?

Daminark

3:10 AM

Then you use Tychonoff's theorem to get that any product of compact sets is compact

Akiva Weinberger

And there's finitely many

Daminark

(o v e r k i l l)

Zach Hauk

well yeah

Akiva Weinberger

So $f(x)<M_1$ on the first set, $f(x)<M_2$ on the second, $f(x)<M_n$ on the last, say

Zach Hauk

yep

Akiva Weinberger

3:11 AM

Since there's only finitely many, we can take the maximum of them.

Zach Hauk

mhm

Akiva Weinberger

So $f(x)<M:=\max(M_1,\dots,M_n)$ on the entire interval, so it's bounded.

Zach Hauk

yep

I understand that

Akiva Weinberger

QED. On $(0,1)$, though, things mess up.

Take $f(x)=1/x$

then the set of open intervals on which it's bounded is a cover has no finite subcover

Daminark

Actually one problem on my pset this week is to prove that if the dimension of a vector space is infinite, the unit ball is necessarily not compact

Akiva Weinberger

3:12 AM

so you can't take the maximum of the bounds because there's always infinitely many, and the maximum becomes infinity.

@Daminark Hm. Maybe do something with the vertices of the $\infty$-octahedron?

The unit points on the axes

Daminark

That's the geometry for sure

But the thing is

Zach Hauk

oh, good point!

Daminark

You don't a priori know you have a Schauder basis

Akiva Weinberger

So no octahedron. :(

Daminark

Well I guess that might not mess it up

It uses a theorem called Hahn-Banach

Akiva Weinberger

3:14 AM

What's a a Schauder basis

Daminark

Basically it's a basis where you can take infinite linear combinations and talk about convergence

Zach Hauk

hmm

i wish my parents would do stuff like that for me, @Akiva

Akiva Weinberger

How do you define the topology, @Daminark

Daminark

So you take a set $\{e_n\}_{n=1}^{\infty}$

Zach Hauk

maybe i sound whiny but i really do.

Akiva Weinberger

3:15 AM

@ZachHauk So Heine-Borel says that a subset of $\Bbb R^n$ is compact iff it's closed and bounded

@ZachHauk Maybe some day you'll be able to visit my house in Brooklyn and I could let you take home a few books

Daminark

Such that for any $x\in X$ ($X$ is your Banach space), $x = \sum_{n=1}^\infty \alpha_ie_i$. This representation is unique

Akiva Weinberger

@ZachHauk Alternatively: Look up some of Ian Stewart's popular math books.

(Popular math meaning math intended for a layperson audience)

Daminark

And actually we're dealing with Banach spaces in particular

Zach Hauk

I was about 20 minutes away from you today (was in the city)

Daminark

So you have a vector space, over $\mathbb{R}$ or $\mathbb{C}$ with a norm

And then $d(x,y) = \|x-y\|$

It's Banach if it's complete under this metric

Akiva Weinberger

3:18 AM

@Daminark I never studied Banach spaces

but they sound interesting

Daminark

I mean I just started recently

I will say, it would've been somewhat nice if we went the more standard route

Which is to do measure theory first

Now we're stuck dealing with $\ell^p$ spaces

While a lot of books do it for $L^p$ spaces

The former is a special case of the latter by considering $\mathbb{N}$ with the counting measure

But yeah, so the proof of the unit ball in infinite dimensional Banach spaces not being compact relies on Hahn-Banach

Zach Hauk

@AkivaWeinberger well if I see you at mathcamp next year...

Daminark

Basically, given some linear functional $f:X\to\mathbb{R}$, where $X$ is a Banach space, you can define its operator norm $\|f\|_{op}$

Zach Hauk

too bad i'm not the best with social situations lol

Daminark

Which is $\sup_{\|x\|\le 1} |f(x)|$

Now, a functional is called bounded if its operator norm is finite

Akiva Weinberger

3:23 AM

@ZachHauk Lol, me neither

Fargle

@Akiva, @Zach: I thought this just came with the territory.

Daminark

So, if you take some closed subspace of a Banach space, say $V\le X$

Akiva Weinberger

Staring at the globe in my fourth grade classroom instead of interacting with my other classmates during recess did make me semidecent at geography, though

Daminark

And you have some functional $f$ defined on some $V$, Hahn-Banach allows you to extend $f$ to all of $X$ while preserving the operator norm.

Akiva Weinberger

Hiding in the stairs instead of playing outside was probably less useful

Daminark

3:24 AM

It's ridiculously overpowered

Zach Hauk

good night

have school tomorrow

Daminark

Alright, see you @Zach

Fargle

@AkivaWeinberger same

Daminark

Lol I'm like, bad at social situations but with a desire to engage

Meaning, I like to talk a whole lot, but my social skills are meh

MickLH

EFFING YES BOY.

Akiva Weinberger

3:33 AM

What

Daminark

Wut @MickLH

MickLH

algorithm breakthrough, 50% resource requirement

Daminark

Dayum

MickLH

and it's the most scarce and important resource that's cut in half

Akiva Weinberger

Woo

What are you algorithming

MickLH

3:34 AM

crypto

Akiva Weinberger

@MickLH Donuts?

MickLH

*2nd most important resource

Daminark

Computational time?

(I'm not sure how much it was using to start with)

MickLH

Actually that's already good, better than factorization schemes at least, it's the size of the ciphertext itself that is reduced

Daminark

Noice

Akiva Weinberger

3:38 AM

SIHT KCARC MHTIROGLA RUOY NAC TUB

MickLH

With a subtle re-arrangement, I can floor divide away a ton of information and reconstruct it on the other side with a best guess, and the corruption is isolated purely into the randomized part, so the message is in-tact

@AkivaWeinberger no but I have another one that can

Akiva Weinberger

WhAt AbOuT tHiS oNe

See, the thing about that one is, whatever tries to read it gets cancer

MickLH

lol

Daminark

1337 sp33k: 1
All the efforts of modern research: 0

6

Also hey @arctic

arctic tern

hey

Simple

3:43 AM

use generating function to show that for a symmetric random walk $P(S_1S_2,\dots,S_{2n}\neq 0)=P(S_{2n}=0)$ for all $n\geq 1$

Maks

Hey @TedShifrin ! Thz for the answer of vector spaces

ALannister

-1

Q: Prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element

I need to prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element, but I'm not really sure how to begin. I know that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ consists of all formal sums $w_{1}[z]_{1} + w_{2}[z]_{2} + \cdots + w_{k}[z]_{k}$ where the $w_{i}$...

arctic tern

@ALannister Any guesses what element might be a generator? (Hint: what's special about the group $\Bbb Z_n$?)

Also, what would you say "ring generated by a single element" means?

Akiva Weinberger

3:58 AM

@DHMO How would one write the "gwoo" sound in Spanish? Güu, perhaps?

Just trying to fill in the sets "ja ge gi jo ju / ga gue gui go gu / gua güe güi guo ??". Could be that it just doesn't exist, though

BAYMAX

5:16 AM

Hi guys..

Any kind of Question/problem sets available for Schwarzs lemma , application of Liouvilles theorem ,open mapping theorem, as these can simply make many complicated looking questions in complex analysis as simple as possible , any site or book for these will be very good!!!...thank you

Daminark

Well, I don't know much offhand about this sort of thing

BAYMAX

means about complex analysis??

@Daminark

Daminark

But I can give general books on complex analysis that I've heard are good

And for which there are pdfs

They'll likely have this result

Stein-Shakarchi, I know is good, that's what they'll use in complex next quarter

Ahlfors is not particularly modern, and also kinda terse, but I've heard is quality

BAYMAX

that would definitely help, .. but i was looking for those which have examples of these concepts as using these results is like using LAW80... :) @Daminark

Thank you !!

Daminark

I mean these are reasonably comprehensive books

So in all likelihood they'll have what you're looking for

BAYMAX

5:28 AM

ok ! got it !!

Actually if one want's to read basic of complex analysis then book by Dennis Zill is awesome ...

@Daminark

Daminark

I'll look into it for sure, thanks!

My path through complex was bizarre, I will need to eventually sit down and learn about this stuff

The bootcamp should have some, done out of Titchmarsh I think

Eric

ewwwwww

BAYMAX

bootcamp out of Titchmarsh ??

Daminark

Partially

Oh @Eric not fond?

Oh lol @Baymax it's this program that my analysis prof holds in the summer

Eric

No I think Titchmarsh is bad and outdated.

I get what he's trying to do by assigning it, but I won't ever be reading it again.

BAYMAX

5:33 AM

Ohh .. Titchmarsh is a programme ?

Daminark

Lolol, do you know why Schlag uses it? When he actually taught complex he used Silverman's translation (likely bastardization) of Markushevich

Eric

It's a book on complex analysis that an analysis professor at our school uses for this summer program he started running last year

Daminark

@Baymax It's a book, it's just that the program I'm hoping to do is going to use it

Eric

I did the program last year, Daminark may do it this year

BAYMAX

Ohh .. both @Eric and @Daminark you are in the same class ??

Daminark

5:35 AM

He's a year ahead of me wrt classes

But we're in the same school and everything

skullpetrol

Which book would you recommend? @Eric

BAYMAX

Ohh..nice..@Daminark ...

Eric

He uses it to "give you technical fluency" which is fair I guess, but it's just so dry and hard to read.

Stein and Shakarchi is really good in my opinion

The entire four volume series in analysis by stein and shakarchi is quite good

That's for basics, beyond that there are so many directions with complex that you can go that no one book is "comprehensive" anymore.

ozarka

How can I show the Cauchy point gives the minimizer of $$m_k=f_k+\mathbf{g_k^Tp}+\frac{1}{2}\mathbf{p^TB_kp}\quad \text{s.t.}\quad \vert\vert p \vert\vert\le\Delta_k$$
along the direction $-\mathbf{g_k}$

BAYMAX

Ok guys..I will definitely look into it ,but there exists no supremum to books or materials i think , hence i am searching .. please you too search as i was telling ..these concepts are bullets .. so if anyone finds any nice material share it ...Thanks...All the best..

ozarka

5:39 AM

Here, the Cauchy point is $$\tau_k =
\begin{cases}
1 &\quad\mathbf{g_k^TB_kg_k}\le0 \\ \min(\frac{\vert\vert\mathbf{g_k}\vert\vert^3}{\mathbf{\Delta_k g_k^T B_k g_k}},1) &\quad \text{otherwise}
\end{cases}$$

BAYMAX

sorry to ask but it belongs to which topic @ozarka

ozarka

This is relating to convex optimization, @BAYMAX.

BAYMAX

Ok@ozarka , thanks!

Daminark

@Eric Are the other books somewhat better at least?

Eric

uhhh well he got rid of the one by Arnol'd right?

I have no idea if Brin and Stuck is good

Daminark

5:48 AM

Fair

Eric

And you can just ask Ted directly if his book is good l o l

Daminark

Oh right yeah, diffgeo

I mean surely

So what does that leave? Probability?

Eric

I would expect Schlag will use it again since it is the perfect length

oh yeah i forgot about that because I just didn't do any of the problems

It was Sinai and I have no idea if it's good or not.

Daminark

Oh speaking of probability, a bunch of people are kinda bummed out that Fefferman's special probability class isn't happening next quarter

At least 2 people I know offhand were gunning at it and it's not being offered

Eric

Oh I didn't know he taught one

Isn't he teaching just regular analysis

Daminark

5:52 AM

Yeah, I think it only so far happened last year

And yes

ozarka

6:05 AM

Would anyone be willing to help me with convex optimization. I am struggling with a problem.

satyatech

7:11 AM

In my book there was a phrase saying that. "x is 2 times less than y " what does it mean,,,man!!

I am confused in this English phrase ,I can't convert it to maths

Hey @Daminark please say mr

Vrouvrou

Hello

Daminark

Hey @satyatech

No need to call me mr, I'm like, probably around your age

So that registers for me as $x = \frac{1}{2}y$

But I'm not terribly fond of that way of saying it

Balarka Sen

7:38 AM

Hi @Daminark, @Alessandro

Hey @Balarka!

Hi @Balarka!

What's up

Not much, I had a lazy weekend. I'm a bit disappointed by some of the oscars

Daminark

Oh those happened?

Balarka Sen

7:43 AM

I didn't see which got them

Alessandro Codenotti

But I'll have an algebraic topology lecture in the afternoon, that's going to be the highlight of the day probably

@Daminark yep, tonight

Ali Caglayan

Eek they mixed up the winner of best picture

Daminark

Huh

Ramanujan

Hello.

Daminark

Damn it's 2, I should get some shuteye lmao

Well, see you guys around!

Balarka Sen

7:54 AM

Night

skullpetrol

8:06 AM

Cya

DHMO

9:10 AM

@AkivaWeinberger One wouldn't. Just leave the gap there.

Compulsive Mathurbator

9:49 AM

So I'm trying to minimise $f(x)=x^8+x^6-x^4-2x^3-x^2-2x+9$ . I differentiated it and factorised it to $f'(x)=2(x-1)(4x^6-4x^5+7x^4-7x^3+5x^2-2x+1)$. I've got no idea how to solve for the other roots or prove absence of any (there are none) apart from numerical methods. Any ideas?

mercio

nope nope nope nope

DHMO

10:35 AM

@CompulsiveMathurbator $f(x) = (x^8-2x^4+1) + (x^6-2x^3+1) + (x^4-2x^2+1) + (x^2-2x+1) + 5$

Compulsive Mathurbator

10:56 AM

@DHMO Thanks

Mateen Ulhaq

@DHMO How do you even notice that?

DHMO

@MateenUlhaq completing the square...

mercio

o_O

DHMO

11:16 AM

@mercio cuando parece que no hay camino, hace una calle

@DanielFischer guten Tag

Akiva Weinberger

11:38 AM

(When it seems there is no way, make a road)

skill patrol

dig a tunnel

user84215

Hello

Akiva Weinberger

@aminliverpool So, out of curiosity,

are you in Liverpool?

user84215

no

user84215

I read somewhere that we can consider the tensor product of modules, say, A and B as dual to Hom(A,B). can you explain it ?

robjohn

12:07 PM

@CompulsiveMathurbator You can use the Sturm Chain to show that the 6th order polynomial has no real roots.

user84215

Speak about the proof of the fact that there exist uncountable non-isomorphic 2-generator groups, and there exist only countably many non-isomorphic finitely presented groups.

Compulsive Mathurbator

@robjohn Thank you

Akiva Weinberger

Interesting anecdote (Alon Amit's answer)

Ramanujan

1:02 PM

hello, Is there anyone who can help me with a geometry explanation?

Compulsive Mathurbator

1:21 PM

Depends on what it is

Mahmoud

How would you solve this ? As efficiently as possible. $$\lim_{x\to \infty} (x-1)^5-(x+1)^5$$

arctic tern

use (x+1)^5 > (x+1)(x-1)^4

or something like that

Mahmoud

You mean $(x+1)(x+1)^4$ ?

Mateen Ulhaq

Or remember how to expand (x-y)^n and do it in your head :-)

arctic tern

@Mahmoud no, (x+1)^5 is equal to (x+1)(x+1)^4

Mahmoud

1:31 PM

Yes but you wrote $(x+1)(x-1)^4$

yes

Huh ?

Factor

if you want efficiency, you want to use an inequality that lets you factor

the inequality will come from x-1<x+1

which implies (x-1)^4 < (x+1)^4

which implies (x+1)(x-1)^4 < (x+1)^5

which implies (x-1)^5 - (x+1)^5 < (x-1)^5 - (x+1)(x-1)^4 = -2(x-1)^4

Akiva Weinberger

The annoying thing about the explanation of the axiom of choice using pairs of socks in infinitely many drawers is that it's possible to describe a total ordering on configurations of sock in 3-space

arctic tern

1:35 PM

but -2(x-1)^4 tends to -infinity

@AkivaWeinberger what?

Mahmoud

So is the right answer, thanks @arctictern

Compulsive Mathurbator

Another option is $a^5-b^5=(a-b)(a^4+a^3b+.... positive terms)$ which give you negative infinity as a-b is negative

Akiva Weinberger

> To give an informal example, for any (even infinite) collection of pairs of shoes, one can pick out the left shoe from each pair to obtain an appropriate selection, but for an infinite collection of pairs of socks (assumed to have no distinguishing features), such a selection can be obtained only by invoking the axiom of choice. - Wikipedia, for example

arctic tern

@CompulsiveMathurbator oh right, odd exponents

@AkivaWeinberger ah, you're saying left and right socks are distinguishable in 3-space.

yes

Akiva Weinberger

@arctictern The one that's closest to the left side of the drawer versus the one that's closest to the right

or, if they're both equally close, whichever one is on top of the other

arctic tern

1:41 PM

...

Ramanujan

Hello

Akiva Weinberger

Modulo some fiddling with details I guess

DHMO

@arctictern no they aren't

Mahmoud

Why are there many definitions of continuity ? And do I need to know more than $\lim_{x\to a}f(x)=f(a)$, to be a good Calculus student ?

DHMO

@Mahmoud you should know the definition of limit also

but the topology definition would not be required for Calculus

Compulsive Mathurbator

1:44 PM

@DHMO my dad likes to pretend they are.

Mahmoud

I do @DHMO. For the topology stuff, I say humbly : See you later !

Akiva Weinberger

Heh, the Mariana Trench has reviews on Google Maps now

One complained that the place lacked atmosphere.

Compulsive Mathurbator

You gotta dig deep to get the dirt on that joint

Akiva Weinberger

(Presumably they've been inspired by xkcd)

Semiclassical

2:00 PM

hi chat

@robjohn Fun fact, I've actually seen Sturm chains used in a physics paper (in order to establish some facts about the spectrum of a particular kind of matrix).

Mahmoud

Hello @Semiclassical $:)$

BAYMAX

Hi all..can any1 help with a physics reasoning question..also it's little awkward to ask in MSE...

Semiclassical

You might as well ask it.

BAYMAX

Ok...thanks ...here it is .. Why the short wave length limit of x-ray produced is independent of nature of target but the characteristic spectra depends on its nature ?

Also can i extend the concept of Bragg's diffraction condition to nanoscale level ? ..

Semiclassical

What's meant by 'nanoscale level'? Keep in mind that Bragg's law is really telling you about the length scale of the underlying periodic structure.

BAYMAX

2:07 PM

I think when $d$ is order of nanometers ?

Semiclassical

Okay.

As a reference point, rock salt has a lattice spacing of about 0.5 nanometers.

BAYMAX

ok , so we must be able to apply Braggs law there , right ??

Semiclassical

Well, take a look here: library.leeds.ac.uk/info/236/learning_resources/268/…

WLB is Bragg himself

BAYMAX

i think the last 3 lines reveals

4-5 lins

Vrouvrou

@DHMO Hello

BAYMAX

2:12 PM

So,i think we can apply

@Semiclassical

Semiclassical

Yeah. So Bragg himself applied his law to the study of x-ray scattering on rock salt.

Mahmoud

This function is very iconic, $$f(x)=\begin{cases} x, \; & x \, \text{rational} \\ 0, \; & x \, \text{irrational} \end{cases}$$ How do we know that : $\lim_{x\to 0} f(x)=0$ ?

BAYMAX

yeah!!

DHMO

@Mahmoud use the definition of limit

@Vrouvrou bonjour

Semiclassical

The following is only a plausibility check. If I consider a sequence of irrationals $\{x_n\}$ converging to 0, then $f(x_n)=0$ so the limit of $\{f(x_n)\}$ is certainly zero.

DHMO

2:15 PM

actually, both $f(x)=x$ and $f(x)=0$ go to $0$

Mahmoud

@DHMO Uhh, how the heck am I supposed to distinguish rationals from irrationals ?

Semiclassical

On the other hand, if I consider a sequence of rationals, I'll get the same since we're taking $x\to 0$.

Right.

DHMO

@Mahmoud I don't see you using the definition anywhere

Mahmoud

What do you mean ?

DHMO

It means, use the definition.

Vrouvrou

2:17 PM

@DHMO je veux revenir sur comment montrer que l'ensemble $D=\{x\in E, d(x,A)<d(x,B)\}$ est ouvert

Mahmoud

@DHMO But near zero there is an uncountable infinity of irrationals, and a countable infinity of rationals ...

DHMO

@Mahmoud I still am not seeing you using the definition

Semiclassical

I imagine the following is moreover true: Suppose $f(x)=g(x)$ on the rationals and $f(x)=h(x)$ on the irrationals, with $g(x)$, $h(x)$ continuous and equal at zero. Then $f(x)$ is also continuous at zero and equals $g(0)=h(0)$.

DHMO

@Vrouvrou oui?

@Semiclassical actually, any partition of the real numbers.

Semiclassical

Makes sense.

I forget, does a partition only involve two sets? ($S$ and $\Bbb R-S$)

robjohn

2:19 PM

@Semiclassical They are useful. I've written some Mathematica code to generate and use Sturm Chains.

DHMO

@Semiclassical I didn't use "partition" in a rigorous way

so it is not a technical term

and it just means $S$ and $\Bbb R \setminus S$

Mahmoud

@DHMO : For all $\epsilon \gt 0$, there exists a $\delta \gt 0$, such that for all $x$ : $|x|\lt \delta \implies |f(x)|\lt \epsilon$

DHMO

@Mahmoud nice

Semiclassical

Ah.

DHMO

@Mahmoud now the $\delta$ is actually quite easy to find, if you think about it

Mahmoud

2:23 PM

Now if $x$ is rational, we can get $|f(x)|\lt \epsilon$ by requiring $\delta=0$

And if $x$ is irrational, we're done because $f(x)$ already equals $0$ for all of them.

Isn't it correct ? D: @DHMO

DHMO

@Mahmoud sorry

@Mahmoud $\delta > 0$ in your definition

Semiclassical

I would imagine that you can generalize to a partition into more than two parts: Start with disjoint sets $\{S_k\}$ such that $\bigcup_k S_k=\Bbb R$. If the restriction of $f(x)$ to each of these domains is continuous and equals $f_0$ at $0$, then so is $f(x)$.

Mahmoud

Oh.

But how is this solvable now ?

DHMO

@Semiclassical I agree

user379685

If I take any point on a circle and draw all the chords that start from this point will all these chords make a disk?

DHMO

2:26 PM

@Mahmoud I trust that you can do it

@user379685 yes

every point on the required disk can be found in one of the chords

SteamyRoot

You definitely can do it for more general partitions

But only the partitions that have $0$ as a limit point matter.

only the sets in the partition*

user379685

@dhmo thanks

Semiclassical

Point. Not much reason to think about a restriction to $x\geq 1$, for instance.

DHMO

@Mahmoud sorry I need to go now

the answer is $\delta = \epsilon$

Semiclassical

If I have a point on the unit circle and then pick a point inside of said circle, that defines a unique line which intersects the circle at some other point. That other point, along with the first one, defines a chord.

DHMO

2:28 PM

I hope that Semiclassical will continue to assist you in my stead

Mahmoud

@DHMO I was writing ..

Semiclassical

Probably not, tbh.

I don't want to do real analysis this morning :/

Mahmoud

Sometimes I just get dump @DHMO, but heh I figured it out, just before you said that you got to go... as strange as it might seem :P

Vrouvrou

@DHMO je reprend la preuve : Soit $x\in D$ on doit trouver $r>0$ tel que $B(x,r)\subset D$ donc soit $y\in B(x,r)$ i.e,. $d(x,y)<r$ et je dois montrer que $d(y,A)<d(y,B)$ on a $d(y,A)\leq d(x,y)+d(x,A)<r+d(x,A)<r+d(x,B)$

BAYMAX

Winners find a way to win !!

Mahmoud

2:31 PM

Bye all ! Time for school.

Thanks for everything !

Vrouvrou

@DHMO vous etes là ?

Nick

There's a poster competition by the Maths Club at my uni tomorrow on the occasion of Science Day. Anyone have any ideas on what I should make my poster about?

i was thinking the Goldbach conjecture but then, I thought can that fill a poster well enough?

Semiclassical

Could do it on conjectures in number theory more generally.

Semiclassical

2:48 PM

@Nick Or you could do a survey of famous problems in number theory, i.e. what they are, a brief statement of history and status.

BAYMAX

3:07 PM

@Nick how about Collatz conjecture ???

Nick

3:26 PM

@Semiclassical sounds like too much put in one A3 poster.