Mathematics - 2018-09-27 (page 2 of 5)

4:00 PM

I'm trying to show that $X$ and $Y$ are convex, and I've recalled the definition that a set $C$ is called convex if for all $x,y\in C$ and $t\in[0,1]$ we have that $(1-t)x+ty\in C$, but I don't know how this translates to sets of sequences.

Alessandro Codenotti

Start thinking about $X$, this one is easier

Oskar Tegby

Okay. Let's just pause here really quick. What I'm trying to do is to see whether or not there exists a hyperplane separating $Z$ and $Y$ where $Z=X-c$. It should've said $Z$ instead of $X$ in my last message. The idea is to use the second geometric form of the Hahn-Banach. It's required that $Z\cap Y=\emptyset$, that they are both nonempty and convex, and that $Y$ is closed and that $Z$ is compact.

Alessandro Codenotti

Well $Z$ is convex iff $X$ is, do you agree?

Oskar Tegby

If this doesn't work, then I only know that this approach doesn't work. If it works, well, then we're done.

I don't know.

mercio

you are going to have a hard time proving that $Z$ is compact

Oskar Tegby

4:06 PM

Hm... $Z=X-c=\{x-c:x\in X\}$

Like, an impossible time?

mercio

yeah

Alessandro Codenotti

So here's an exercise for you to get familiar with convex sets: Prove that the translation of a convex set is convex

Oskar Tegby

Okay. Visually, it's obvious. Just a small comment: As $Z$ isn't compact, then maybe it's closed and I can show that $Y$ is compact instead.

mercio

Oo

oh you meant for the convexity ?

Oskar Tegby

Yeah

mercio

4:08 PM

you are going to have an equally hard time proving that $Y$ is compact

Oskar Tegby

Okay. Then, we can't use Hahn-Banach. Right?

mercio

I don't think so

Akiva Weinberger

A convex hull is the intersection of all half-planes containing the shape, yeah?

Alessandro Codenotti

Hmmm I'm not sure those two sets are actually separated by a closed hyperplane

mercio

I don't think they are

Oskar Tegby

4:09 PM

Okay.

It says that I should compare it to problem 1.9. Thus, here it is.

mercio

maybe exercise 1.9's conclusion is false in spaces that are infinite dimensional

$X$ and $Y$ sure seem to be non empty disjoint convex sets

Secret

Finally, after 4 hours of work, dynamic zoom is finally working

$$width*(0+(0-u)*(v-1)+(\frac{1}{2}-u))$$

Who said linear algebra is easy

It is NOT

rschwieb

@Secret what was hard?

Daminark

Did someone say functional analysis?

Prashin Jeevaganth

Hi testing, I just found out about this chat feature in MSE...

Tobias Kildetoft

4:16 PM

@Daminark No. Nobody said that. No need to stay...

Akiva Weinberger

Did someone say disco?!

Daminark

Oh :(

Secret

@rschwieb Linear algebra is not as trivial if you want to move something to the center of the screen, and then do a zoom there

Daminark

But I still sense Brezis

Akiva Weinberger

You programming? @Secret

rschwieb

4:16 PM

@Secret What framework are you using?

Secret

Processing (java basically)

rschwieb

doesn't sound hard!

Well, if I had to do it in java it would be hard

Secret

@AkivaWeinberger Yeah, programming the number plotter that plots everything you specify between [0,1]

rschwieb

@Secret It's a 3-d model?

What is there beyond basic projective transformations?

Oskar Tegby

You're right, @Daminark. Functional analysis is, what some say, dope.

mercio

4:19 PM

I thought that X and Y can't be separated would be an easy consequence of adh(X+Y)=E and X+Y <> E, but I'm not sure anymore

s.harp

@Secret that sounds like affine or projective geometry, not linear algebra 8)

Oskar Tegby

Why wouldn't be true in infinite dimensions?

mercio

why would it be true in infinite dimensions ?

Secret

Ah, no wonder it is so hard when all I want is to shift the origin and then double the size of the object at the new origin

Oskar Tegby

Because it's true in finite dimension. I mean something along the lines of what role that the finiteness has for this phenomenon. It's at least an interesting question.

mercio

4:22 PM

LOL

rschwieb

@Secret A year or two ago i remember programming a dolly zoom transformation, and AR projection onto a video. That was fun

Oskar Tegby

What?

What's fun?

Daminark

In infinite dimensions you need to throw topology in there. I think you want that one set is closed and another compact in order to guarantee it

Oskar Tegby

Alright

mercio

it was fun because you didn't add anything to the "it's true in finite dimension so it must be true in infinite dimensions" view that you had expressed just earlier

and anyway you better drop that line of thinking

Secret

4:24 PM

Demonstration

Oskar Tegby

Okay.

s.harp

This is a static image, right

mercio

and I am going to be that guy that spent 15 minutes thinking deeply and say in the end "yes, it is obvious, from Adh(X+Y) = E and X+Y <> E (and X Y are subvector spaces) that X-c and Y can't be separated"

Oskar Tegby

What's $Adh(\cdot)$?

mercio

french for closure

Mary Star

4:25 PM

Hello!!

We have the sentence:
$\forall x\in \emptyset \ \exists ! y\in \emptyset : (x,y)\in \emptyset $

This is true because $\forall x\in \emptyset$ is aleays true and so the whole sentence is true, or not?

But the sentence:
$\forall x\in \{1,2,3\} \ \exists ! y\in \emptyset : (x,y)\in \emptyset $
is false, or not? But why? Becayse when we take for example $x=1$ then we have that $\exists ! y\in \emptyset$which cannot be true?

Secret

it is static because uh, I don't know how to make a gif in 3 seconds

Oskar Tegby

What does $X+Y<>E$ mean?

mercio

it's computer science for not equal

Oskar Tegby

Okay.

Why is it obvious?

mercio

because I thought a lot about it

Oskar Tegby

4:27 PM

I wish I could write that on my exams.

s.harp

I'm trying to think of examples where infinite dimensional spaces are strongly different to "obvious" finite dimensional intuition

for example the unit sphere in most standard infinite dimenisonal spaces is contractible

Alessandro Codenotti

@s.harp Compactness of the closed unit ball is the go to example

Oskar Tegby

High dimensions are weird.

s.harp

@AlessandroCodenotti I don't like that example I have to confess

Oskar Tegby

Is it obvious to anyone else than mercio why there isn't any hyperplane that separate them in infinite dimension?

Secret

4:32 PM

Now I can finally deep dive into the 4! decimal place without too much chore

Alessandro Codenotti

@s.harp Why? It's actually a characterization

(but that's not trivial)

mercio

@OskarTegby also if you know what the dual of $\ell^1$ is you can try to build yourself a continuous linear functional that separates $X-c$ and $Y$ and then fail at that

s.harp

@AlessandroCodenotti because I don't really find it that contradictory to intuition that in an infinite dimensional space you can move in infinitely many directions

That unit balls are very often weakly compact but sometimes not seems different to intuition, the weak topology in a way cuts off large parts of "infinity", but sometimes not large enough parts

Soham Chowdhury

@Secret schizophrenic?

Oskar Tegby

That's a bit harsh.

Secret

4:39 PM

Schizophrenic number

A schizophrenic number (also known as mock rational number) is an irrational number that displays certain characteristics of rational numbers. == Definition == The Universal Book of Mathematics defines "schizophrenic number" as: An informal name for an irrational number that displays such persistent patterns in its decimal expansion, that it has the appearance of a rational number. A schizophrenic number can be obtained as follows. For any positive integer n let f(n) denote the integer given by the recurrence f(n) = 10 f(n − 1) + n with the initial value f(0) = 0. Thus, f(1) = 1, f(2) = 12, f...

Oskar Tegby

Haha! Oh! Okay.

Nicolas

Lovely: Volume II of Yaglom's "Correlation Theory of Stationary and Related Random Functions" is entirely footnotes to Volume I.

s.harp

@Nicolas was it written at the same time or a few years later?

Oskar Tegby

What would even the dual of $\ell^1$ be?

mercio

it is the vector space of continuous linear functionals on $\ell^1$

s.harp

4:43 PM

Do you know what the dual to $\ell^p$ is for $p\neq1$?

Nicolas

@s.harp At the same time I believe. The main text also refers to the notes a lot. He has basically spun half of the work out into notes.

s.harp

@Nicolas the other option would have been too crazy

mercio

nonzero continuous linear functionals correspond to closed hyperplanes so I thought you would know

also exercise 1.9 talks about $E^*$, so really you should know what a dual space is

but if you haven't covered the dual of $\ell^1$ in the notes, it might be time for another exercise about figuring it out

Oskar Tegby

Okay. I just feel like it's a lot of verbal motivation, and not so much theoretical proofs.

Maybe that's what advanced mathematics is like.

mercio

I am just leaving the theoretical proof to you because it is you who are doing the exercises and not us

rschwieb

4:50 PM

In advanced mathematics, verbal motivation helps you remember how the theoretical proofs work.

3

In physics, verbal motivation helps you distract yourself from the fact that your definitions are soft and your mathematics are questionable.

3

Secret

5:02 PM

Ok, I ran out of precision. This is the deepest zoom obtained so far for the liuoville number 0.320005000000000001

Prashin Jeevaganth

Secret

In reality, the same pattern of 10 sticks repeats for each stick countably many levels deep

thus ensuring uncountability of liuoville numbers

Prashin Jeevaganth

Can anyone tell me why the answer is as such? Or is this place the wrong platform to ask exam questions?

Rithaniel

The proposition is that if ab is even, then a or the b is also even. The contrapositive is that if both are odd, then the product is odd. That was shown to be true, so it's contrapositive is also true.

Prashin Jeevaganth

Is this OR statement at the conclusion(RHS) an XOR or inclusive OR?

I'm confused with the word either, like what are instances of XOR in English and what are instances of inclusive OR in English?

Rithaniel

5:09 PM

Well, an even times an even is also even.

Inclusive OR, though, is usually the meaning of the word "or"

XOR should require an additional clause just as "but not both"

Prashin Jeevaganth

So is it safe to say "either" holds no weight here?

Rithaniel

A person might be sloppy with their phrasing, however, and say "or" when they mean XOR.

Prashin Jeevaganth

@Rithaniel Damn, if this happens, do I have to determine what's fight with my logic?

Rithaniel

Well, if neither is even, then it's equivalent to saying that a is not even and b is not even.

Go by what is stated in the text, I'd say. If a person is sloppy with their wording, that's their failing.

Will Njundong

When they say:

Find a formula for F(x) = http://i.imgur.com/9wCGj6U.png
and then find F'(x). What do you notice?

I'm unsure what im supposed to be looking out for

Rithaniel

5:15 PM

Well, evaluate the integral over t, and then take the derivative with respect to x

Will Njundong

So get the anti deriv, replace t with x, and figure out the derivative?

Rithaniel

don't forget to also subtract the anti-derivative where t=0. It won't affect your final answer, but it's good to remember details like that.

Will Njundong

good point, thank you

Rithaniel

No problem.

Will Njundong

so i end up with the same equation but with x'es instead

Not sure what thats supposed to tell me, tbh

Rithaniel

5:20 PM

Well, it seems that you've already noticed the thing they might be trying to show you.

Will Njundong

-_- gosh i would fail terribly if i was a math major. Is this what those guys spend their time doing?

I often cant understand or am not interested enough in figuring out the "why" behind how these things are done in math.

Trying to think of where that is applicable and im just blank

rschwieb

@Semiclassical Did you ever consider using "Pseudoclassical" or "Quasiclassical"? or Hemiclassical?

Semiclassical

no. the first and last aren't actual terms people use

and semiclassical is the more typical word than quasiclassical

rschwieb

@Semiclassical How about 3/4classical

\sqrt{classical}

oo, oo, Ultraclassical

so many prefixes to choose from

Semiclassical

yeah, uh, have fun with that

rschwieb

5:25 PM

@Semiclassical Sorry for the flippant streak. It's already over. Looks like it didn't appeal to you.

Secret

Ok, I am going to call this a day. In order to dive deeper, I will need to code in hardcore java to take advantage of the BigDecimal datatype

Rithaniel

There's a whole list of latin prefixes on wikipedia

rschwieb

I think maybe I shoudl have used rschw, since that ends right before the vowels everyone transposes

Secret

But before closing this chapter, some pretty stuff:

A certain class of $G_{\delta}$ set formed by arbitrary intersection of punctured open intervals will look something like the black regions combined for the orange plot

Ted Shifrin

Hi, DogAteMy.

Secret

5:28 PM

Irrationals and liuoville numbers belong to this class

Ted Shifrin

or Liouville :)

Semiclassical

hi @ted

Ted Shifrin

Liouville numbers are transcendental, hence irrational. So I don't get your sentence.

hi @Semiclassic

Secret

As to be elaborated later, the black band is to be understood as the series of convergining black bands stuffed within to fill in all space, and this procedure is repeated countably many times. Thus each line in the black band actually has uncountably many numbers in it, regardless of how deep the zoom is

Ted Shifrin

Do all irrationals really have whatever property you're talking about?

Secret

5:30 PM

proofwiki.org/wiki/Irrational_Numbers_form_G-Delta_Set_in_Reals

Well, both the set of irrationals and its subset the liuoville numbers are $G_{\delta}$ sets, meaning they can made by arbitrary intersections of open sets

Ted Shifrin

Oh, I see ... context matters.

Secret

And in the above plot up to denominator of 400, the white plots are all the rationals up to that denominator

Thus whatever is left behind are irrationals, and the exact same pattern is seen when going higher or lower in denominators

Prashin Jeevaganth

Can anyone explain the answer? Who should we start with first?

I hate the names Aikan and Dueet cuz not(AikanDueet)

Soham Chowdhury

Well, Kenyu tell us if you've done any similar problems, perhaps simpler ones?

Secret

Every black band has the following basic structure. Within each ,you also see countably many black bands, each of these have the exact same structure shown

Prashin Jeevaganth

5:35 PM

I have another past year question, I solved it by luck, now I can't solve this

Secret

and this pattern continues countably many levels deep

Ted Shifrin

@Prashin: Well, let's start here. Could Aiken be lying?

Prashin Jeevaganth

Think not, theres a contradiction

Ted Shifrin

Did you work that out?

Prashin Jeevaganth

wait

I'm not too sure if I'm thinking the right way

Ted Shifrin

5:37 PM

Well, if Aiken is lying, he must be the knave or the spy.

What about D and K?

Oskar Tegby

@Ted: I'm stuck. :/

Ted Shifrin

How do you do, stuck? :) Howdy @Oskar

Rithaniel

First important detail is that at least one of the three statements has to be false.

Ted Shifrin

OK, @Rithaniel. Let Prashin work on it.

Secret

Meanwhile, the rationals are $F_{\sigma}$ sets, formed by countable union of closed sets. Its structure is "pseudo" self similar as plotted

Rithaniel

5:38 PM

(Gotcha)

Oskar Tegby

Prashin Jeevaganth

Can K be a spy?

Ted Shifrin

Oh, you finished #1, @Oskar? Cool.

Oskar Tegby

I have these parts left.

Prashin Jeevaganth

damn should a spy say that he's a spy? I don't know what's the use of the spy

Ted Shifrin

5:40 PM

What about D?

Rithaniel

Calculating the decimals of an irrational square root by hand. What is the method?

Secret

babylonic method

Ted Shifrin

Ah, there's an algorithm for that. Surely it's on the web, @Rithaniel. We all learned that in school when I was a kid.

Prashin Jeevaganth

That leaves that D is telling the truth so there's a contradiction right cuz we assumed that he's a liar

Oskar Tegby

Barbaric method

Ted Shifrin

5:41 PM

@Prashin: No, I don't follow. If A is a liar, then D has to be lying.

@Oskar: What is $c$?

Secret

Oskar Tegby

$c_{2n-1}=0$ and $c_{2n}=\frac{1}{2^n}$ for all $n\in\Bbb{N}$.

Secret

More details later in another room

Now to go to sleep

Rithaniel

Fair enough. To google then. (I've recently stopped going to google for answers to questions I think might be obscure. Sometimes you just can't get the data you want and it's faster to just ask someone.)

Prashin Jeevaganth

Hmm, ok so now there are 2 liars, how can K be a spy?

Ted Shifrin

5:44 PM

@Prashin: You tell me. Can he?

@Oscar: So why is $Y\cap Z=\emptyset$?

Alessandro Codenotti

Hi @Ted

Ted Shifrin

hi @Alessandro

Prashin Jeevaganth

@TedShifrin ok it seems that a spy cannot say that he's a spy since a spy can possibly lie which leads to a contradiction? With that he's either a knight or knave, but he's statement says he's a spy so he must be lying?

Ted Shifrin

Oops, @Oskar (sorry about typo): So why is $Y\cap Z = \emptyset$?

I don't think you understand the contradiction, @Prashin.

Semiclassical

w

oops

Oskar Tegby

5:49 PM

Here, $Z:=X-c=\{x-c:x\in X\}=\{x=(x_n)_{n\in\Bbb{N}}\in E:x_{2n}=-\frac{1}{2^n}\forall n\in\Bbb{N}\}$. If $x-c\in Y$, then we have that $y_{2n}=\frac{1}{2^n}y_{2n-1}=-\frac{1}{2^n}$ for all $n\in\Bbb{N}$, which means that $y_{2n-1}=-1$ for all $n\in\Bbb{N}$, but that would mean that it would diverge as $y_{2n}=-\frac{1}{2^n}\to0$ as $n\to\infty$ and $y_{2n-1}\to-1$ as $n\to\infty$. Right?

Semiclassical

What I'd go through to start: Which of them can possibly be the knight?

Ted Shifrin

Is that what they mean, @Oskar?

Perhaps so. I was thinking of $X-\{c\}$.

Oskar Tegby

Well, if $x-c\notin Y$, then $Z\cap Y=\emptyset$, as desired.

I know something at least. Not as slow today. :)

Ted Shifrin

We're not talking about convergent sequences, @Oskar.

Oskar Tegby

Oh! Sadness. We're not? :/

Prashin Jeevaganth

5:52 PM

@Semiclassical Only Aiken can be a knight since if Dueet is a knight, there's a contradiction since there must be no other knights other than himself. For Kenyu, a knight can't possibly be a spy

Ted Shifrin

OK, I withdraw from the puzzle discussion.

Semiclassical

Right.

So the question is which of the two remaining is the spy and which is the knave

Ted Shifrin

I don't understand what you're saying $Z$ is, @Oskar.

Oskar Tegby

$Z=X-c=\{x-c:x\in X\}$

Semiclassical

What I'd now focus on is Dueet's statement

Ted Shifrin

5:53 PM

But $c\in X$, isn't it?

Oh, no, $X$ had all even terms $0$.

Oskar Tegby

No? We have $c_{2n}=\frac{1}{2^n}$, and we require $x_{2n}=0$.

Tanner Swett

What's the standard term for "a rectangle, including the boundary of the rectangle and all of the points inside of the rectangle"?

Like, what is that point set called?

Ted Shifrin

Right, I see. So all the even terms in an element of $X-C$ are indeed $-1/2^n$, and so if this point is in $Y$, we have $y_{2n}=-1/2^n$ and $y_{2n-1} = -1$, as you said. OK, cool. Now what's the problem? What's our big Banach space? It's not the space of convergent sequences.

Tanner Swett

The analogous point set for a circle is called a closed circular ball.

Semiclassical

I feel like what you're really asking is whether a rectangle is defined to include its boundary or not.

ÍgjøgnumMeg

5:55 PM

A box?

Oskar Tegby

We have $E=\ell^1$.

Ted Shifrin

I would call it the closed rectangle, as one can also have the open rectangle called a rectangle.

So is $y\in E$? @Oskar

Prashin Jeevaganth

@Semiclassical I guess Dueet has to be spy since being a knave forces him to lie about Aiken, but there's a contradiction... so Kenyu must be a knave instead

Tanner Swett

@ÍgjøgnumMeg That's what I thought, but I don't know of any authors that have actually used the word "box" that way.

Oskar Tegby

No? @Ted

Semiclassical

5:56 PM

Right. in which case Kenyu must be lying when he says he's the spy

Ted Shifrin

But not for the reason you said, @Oskar.

Semiclassical

which is perfectly fine, since he's in fact the knave

So the claimed answer is indeed correct.

Oskar Tegby

Hm... @Ted

Tanner Swett

@TedShifrin I thought about something like that, but calling it a "closed rectangle" seems a little confusing, since the boundary of the rectangle is also a closed set.

Ted Shifrin

Yeah, but mathematicians (beyond elementary school) don't refer to rectangle as the boundary, typically.

To be on the safe side, define your term.

Tanner Swett

5:58 PM

Hmmm, all right.

ÍgjøgnumMeg

the closed angular disc

Tanner Swett

Thanks.

I could call it a "closed rectangular ball", but that sounds weird. :D

Semiclassical