Mathematics - 2016-04-28 (page 1 of 2)

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12:50 AM

@Agawa001, I think that was from the movie Scarface, "First you get the money, then you get the women."

Hello, can anyone answer the question?

2

Q: Why do we classify infinites in so many symbols and ideas?

I recently watched a video about different infinities. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \ldots, \omega_\omega$, etc.. I can't find myself in all of this. Why there are so many infinities,...

1:05 AM

many of those are ordinals, not cardinals

anyway you can just look up "ordinal" and "cardinal number" on wikipedia and it'll explain what they are

if you know what they are then it should follow you know how they're different

I don't actually know what $\varepsilon_0$ means either.

@anon are you familiar with cyclic codes ?

no

is anyone familiar with cyclic code

Someone somewhere is...

:-D

1:39 AM

yo

@MikeMiller Whoa. Way wrong. Isometric embedding (even locally) is much harder to prove than Whitney. Yes, and dimensions have to be very high. This is serious differential systems stuff.

@TedShifrin

are you familiar with cyclic code

???

I know nothing about codes, Karim.

Stop spamming the room.

ok ok

anyone know Lipschitz conditions?

1:42 AM

@Ted: that's not what I said.

It's hard to read carefully with so many things going on ...

but it would be nice if you did before you called me wrong :)

It looked to me like it was stated as obvious that every Riemannian manifold could be isometrically embedded.

Well, I read it through twice.

A function $f :(a,b) \rightarrow R$ satisfies a Lipschitz condition at $x \in (a,b)$ iff there is $ M >0$ and $ \epsilon >0$ such that $ \mid x-y \mid < \epsilon$ and $y \in (a,b)$ imply that $ \mid f(x) - f(y) \mid < M \mid x-y \mid$. Give an example of a function that fails to satisfy a Lipschitz condition at a point of continuity. If f is differentiable at x, prove that f satisfies a Lipschitz condition at x. \\

By the Mean Value Theorem, there exists a c between x and y such that $x,y \in (a,b)$ and \\

sure. I agree that's nowhere near obvious or trivial or anything like that. The conversation had transitioned strictly to amooth manifolds, by my reading when I said that, before I said so.

1:44 AM

@MikeMiller Starting here and reading about 5 lines.

Anyhow, I should just not bother unless I'm here, and even then maybe I shouldn't bother.

Why is the pullback of a non-vanishing $\alpha^*(\omega)$ of a $k$-form $\omega$ on $\mathbb{R}^n$ also non-vanishing ($\alpha$ being a coordinate chart for a $k$-manifold)?

That needn't be correct, @Simeon.

@Ted: See the last parenthetical.

maybe f'(c) is the M portion of this :S

That's what I thought, but I thought I read it somewhere

1:46 AM

Pull back the non-vanishing $x\,dx + y\,dy$ on $\Bbb R^2-\{0\}$ to the unit circle and you get the $0$-form.

because it exists and it's bounded by (a,b)

Oh sorry, I meant that if $\omega$ is vanishing $\alpha^*(\omega)$ is vanishing

Is that really what you meant? Pullback is linear.

I doubt that's what you meant, @Simeon.

Oh never mind, that's obvious

1:48 AM

What you said originally, with the important parenthetical you had at the end, is correct.

Well, @MikeM, the original question didn't actually make sense.

Think about 1) what it means to be a coordinate chart 2) the definition of pullback.

Sure.

Unless you mean we're pulling back by another diffeomorphism $\beta$ which didn't appear.

I can see how it's true in the case when $\omega = dy_I$ (some elementary $k$-form on $\mathbb{R}^n$), but what about the general case?

I assume what was meant was "if $\alpha$ is a coordinate chart, and $\omega$ is nowhere vanishing, why is $\alpha^*\omega$ nowhere vanishing?" That's what I was trying to answer.

1:50 AM

Yes

but $\omega$ is a $k$-form on $\Bbb R^n$, so its being nowhere vanishing doesn't mean restriction will be nowhere vanishing. That was precisely my point.

So are we talking about a nowhere-vanishing $\omega$ on $\Bbb R^n$? Or are we already saying it's nowhere vanishing when restricted to the submanifold, which probably isn't defined yet?

It's restricted to a manifold

We are not restricting anything to a submanifold. Coordinate chart. We're pulling back by a diffeomorphism.

Then my counterexample stands.

The $1$-form $x\,dx + y\,dy$ is nowhere zero on $\Bbb R^2-\{0\}$, but pulling back by a parametrization of the unit circle we get the 0-form.

Can someone explain to me what cardinality of $\mathbb{R}$ means?

1:54 AM

$\omega$ is defined on a $k$-manifold in $\mathbb{R}^n$

You said you had a $k$-form on $\Bbb R^n$, @Simeon. It is really important to get the question specifically right :P

Do you even know what it means to have a form on a manifold?

@Ted: It seems hard to call projecting to the unit circle, which is not a diffeomorphism, a "coordinate chart".

But I'll stay out of this.

@MikeM: The original question said $\alpha$ was a chart for a $k$-manifold.

I'm sorry, I didn't word my question properly, the form $\omega$ is defined on the $k$-manifold.

I'm not projecting anybody anywhere.

I'm outta here.

2:01 AM

evening chat

Can someone explain what cardinality of $\mathbb{R}$ means???????????????

2:13 AM

isn't cardinality the size of a set?

yeah it is the size of the set. so it's like how many elements are in the set of real numbers

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

does the question of "how many" make sense with infinitely many?

oh yeah x.x!

2:55 AM

hi

@JesterTran a cardinal number is essentially a measure of the size of a set. it is a generalization of counting numbers, which measure the size of finite sets.

@anon how do you "measure" the size of the reals when it's uncountable?

we say two sets have the same size if they are in one-to-one correspondence. we say one set is smaller than another if the first fits inside the second but not vice-versa.

this is all written about to death a million places.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

3:21 AM

@JesterTran have a look at this

3:55 AM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ thanks

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

4:08 AM

thanks for asking :-)

1 hour later…

5:23 AM

Where do I post to get technical support with Stack Exchange related issues?

I can't @ people in comments anymore. Was this a mod action against me for being too sassy or is it a bug I need to remedy?

hi @Axoren are you familiar with coding theory ?

There are so many things that could mean. I believe you asked me once and I don't remember what you meant before.

Programming code? Or like Huffman coding?

Or like Computation theory?

like reed solomon code etc

error correcting codes etc

Ahh, that stuff always looked fun but requires way more information theory than I have under my belt.

oh i see

5:31 AM

am i just being thick, or is the question seemingly being asked for in comments in direct contradiction with the question asked?

@Semiclassical maybe you could help me understand something in the notation

of coding theory? i'll pass

no in algebra

such as?

Note to self: Never open with anything related to coding theory. People will run.

More of a reason to keep my belt loose.

5:34 AM

i vaguely remember hamming metric stuff from abstract algebra, but that was years ago

and i really haven't touched finite fields

$F_2^3 = F(\zeta)$ so we have N = 010011100 is 9 bit string why is that represented as $m(x) = \zeta*x^2 + (\zeta + 1)x + x^2$ ?

pretty sure that counts as a coding theory question :/

no

it is finite fields

you're asking why a 9-bit string is represented a certain way in a finite field.

to say that's not coding theory is a bit disingenuous.

It's almost like lying, though. I don't know many times Finite Fields are used outside of cryptography.

@Adeek What is $m$ here?

And that big ol' $F$

5:38 AM

m = 3

I hope not. $3(x)$ is a function I already have trouble fathoming.

are you sure that the last term in the $m(x)$ you quoted is $x^2$?

no

that is zeta^2

okay, now i think i can see it

$\zeta x^2 + (\zeta + 1)x + \zeta^2$

why ?

5:40 AM

your 9-bit string can be chopped up as 010 011 100

ok

the way i read the first three bits is that "the leading coefficient does not have zeta^2, does have zeta^1, and doesn't have zeta^0"

oh I see

yes that makes sense

@Semiclassical thank you

same interpretation works for the other triplets

yes

is it possible to do finite field arithmetic with wolf ram alpha ?

5:50 AM

well, Mathematica supports finite field arithmetic as a package that has to be loaded rather than being present by default

oh

which would make me guess that wolfram alpha won't have it natively

on the other hand: wolframalpha.com/examples/FiniteFields.html

oh awesome I need that

that gives you background on a given field, if not a way to do arithmetic directly

hm

here do we consider +- = B1 as what ?

5:53 AM

and now you're entirely past me

as 1 symbol ?

ok nvm

I will figure it out

I need to finish this assignment today I hate long assignment

it sucks

this looks handy: ascii-code.com

and includes the ref for the plus-minus sign

oh awesomeee

I thought that +- = B1 was like more than 1 character

I need sleep lol

yeah this question isn't bad

@Semiclassical I want to divide $x^4 + \zeta^2x^5 + x^6(1 + \zeta^2)$ by $(x + \zeta)(x + \zeta^2)(x + \zeta^3)(x + \zeta^4)$ and obtain the remainder can wolf ram alpha do that ?

within the field? no clue

6:09 AM

yeah

6:32 AM

hey @TobiasKildetoft

Tobias Kildetoft

@Adeek Hi

is there a fast way of computing the remainder when I divide $x^4 + \zeta^2 x^5 + x^6*(1 + \zeta^2)$ by $(x + \zeta)(x + \zeta^2)(x + \zeta^3)(x + \zeta^4)$ I want to obtain the remainder

after the division

Tobias Kildetoft

@Adeek what is $\zeta$?

so we are considering $GF(2^3) = F_2(\zeta)$ where $\zeta^3 = \zeta + 1$

Tobias Kildetoft

Not sure about a fast way.

6:37 AM

do I have to expand g and then do long division :S ?

Tobias Kildetoft

@Adeek That would work at least, and I don't know another way

why do you want to find the remainder?

I need it for a code

I need it to compute a specific code

do you have some algebraic system @TobiasKildetoft ?

Tobias Kildetoft

@Adeek some what?

like mathematica

Tobias Kildetoft

Ahh, not really. I mainly use GAP since that can usually do what I need to do

6:40 AM

can it do those operations ?

I just wanted the remainder instead of doing 1 hr computation

Tobias Kildetoft

no idea. It is meant for group theory stuff

probably sage can do it

oh

I really hate mindless calculation assignments like this

it sucks

Akiva Weinberger

7:32 AM

@Axoren I would post it in either math meta or stack exchange meta

Mathematics Meta

Meta Stack Exchange

sounds like a question for meta with a "bug" tag

as far as it being an action done by a mod; I would try posting the question in the math mods' office

8:25 AM

Hello there! I have a simple question, can an operator operate on numbers or is it only for functions?

does a mathematical history paper require an abstract?

proof papers have

Abstract
Introduction
blah

Tobias Kildetoft

@usukidoll all papers should have an abstract which states what the paper is about

Why would it only be functions? @Aldon

6

Q: Approximate spectral decomposition

See attempt below I am interested in effective and constructive computations for finding approximate spectral decompositions in some suitable format. Namely, let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\},...

linear-algebra hilbert-spaces approximation computability constructive-mathematics

Someone to comment?

Tobias Kildetoft

@Aldon the term operator is not even quite universally defined (meaning that not everyone will agree what the term means)

8:36 AM

:S

1/x is uniformly continuous on (0,1)... what the heck? there's a bunch of not uniformly continuous proofs out there

hello

does anyone know how conditional expectations can be used in hypothesis testing?

what role do they play?

Tobias Kildetoft

@StanShunpike It seems that you got an answer to your question about this two hours ago

Yeah, still confused

Tobias Kildetoft

not mentioning that here seems like it might just cause people to repeat stuff you have already heard

He didn't define conditional expctation

he discussed conditional probability

like

i know what conditional expectation is

i just don't understand why it would come up in this contex

8:43 AM

@skillpatrol I thought they only work on operators (I always thought of the gradient operator as an example). So it could work with scalars and vectors too right?

yes @Aldon

Tobias Kildetoft

@StanShunpike Sure, I am not saying the answer is complete (I did not even read it). But you need to make people aware of what information you have already gotten on the topic so they don't waste their time

is $\frac{-1}{2\sqrt{x}}$ unbounded at f:(0,1) ->R

@TobiasKildetoft Hmm...good point

I guess I could have been more specific in my question

thank for the feedback

Tobias Kildetoft

@StanShunpike So just link your question along with what you ask here

(and of course possibly explain what you are still confused about)

9:23 AM

Welcome @USER91500

@usukidoll Draw something, it should be relatively easy to see if something is unbounded or not.

I did

So I guess you have answered your question?

9:39 AM

Hi guys! What are the physical applications of the Tree Matrix Theorem of Graph Theory?

Tobias Kildetoft

10:12 AM

@FrancescoS Why do you expect it to have any?

@TobiasKildetoft because my prof asked me this question.

Suppose $A = (-1,1)$, how do we find a nonconvergent sequence in $A$? (convergence in $\mathbb{R}$)

Tobias Kildetoft

@JesterTran just alternate between two values

Hi @AkivaWeinberger.

Akiva Weinberger

Does anything interesting happen if we make a chain complex out of pointed topological spaces rather than abelian groups? Like, $\xrightarrow\partial C_n\xrightarrow\partial C_{n-1}\xrightarrow\partial$, where $\partial^2$ is equal to the constant map that sends everything to the base point $*$, and we define the homology spaces as $\partial_n^{-1}(*)/\partial_{n+1}(C_{n+1})$.

10:19 AM

@TobiasKildetoft $\{0,1,0,1,0,1,\dots\}$?

Akiva Weinberger

@BalarkaSen Hi

Tobias Kildetoft

@JesterTran $1$ is not an element in that set

So $C_n$ here are based topological spaces?

@TobiasKildetoft replace 1 with $\frac{1}{2}$, sorry

Tobias Kildetoft

@JesterTran then yeah, that works

10:20 AM

And what does $\partial$ do?

Akiva Weinberger

@BalarkaSen Yeah. Could be that it doesn't turn out to be interesting

Tobias Kildetoft

@AkivaWeinberger what sort of object are those quotients?

pointed sets?

Topological spaces, I guess.

Those are quotient spaces.

Tobias Kildetoft

sounds like a terrible idea then. The whole point of homology is to get objects of a type we understand better

Akiva Weinberger

Pointed topological spaces. We contract the image of $C_{n+1}$ to a point.

10:21 AM

@TobiasKildetoft thanks, how about $\left\{ \frac{\sin k}{k}\right\}_{k=1}^{\infty}$?

Not really sure if this is interesting either.

Tobias Kildetoft

@JesterTran that converges to $0$

Akiva Weinberger

I mean, I haven't really thought about it that much; I just realized that it was define-able

Like, the basic definition of homology works here

@TobiasKildetoft how about if $A = (0,1)$, is that now a nonconvergent sequence in $A$? (convergence in $\mathbb{R}$)

I mean trying to do homological algebra in Top is probably going to fail miserably as it's not an abelian category.

Tobias Kildetoft

10:23 AM

@JesterTran sure, it converges to something outside $A$, but it no longer has values inside $A$

Akiva Weinberger

What's the category of pointed top. spaces called? (Also, I'm not 100% sure what an abelian category is, 'cause the definition seemed complicated)

Top_*

Tobias Kildetoft

@AkivaWeinberger an abelian category is just a category that is pretty much in every way like the category of modules over some ring

That ^

Tobias Kildetoft

In some sense it is exactly that due to the embedding theorem (with some minor caveats that I can never recall)

10:25 AM

Right, I remember someone in here telling me that.

Probably archipelago.

@TobiasKildetoft Ok, am I understanding convergence in $\mathbb{R}$ correctly: a nonconvergent sequence in an open interval (say, $A$) is a sequence that is a subset of $A$ and does not converge in $A$?

Akiva Weinberger

Eh. Arright, thanks, guys

Tobias Kildetoft

@JesterTran yes

@TobiasKildetoft thank you very much!

@AkivaWeinberger Anyway, what Tobias said is more down to earth: what is this going to accomplish? Homology helps to study topological spaces and it's homeomorphism (well, homotopy) type by studying groups associated to them in a functorial way. What's the point of your construction? What is it going to help me study?

Defining things out of blue is not interesting in itself.

Oh cruds I forgot I have work to do.

Akiva Weinberger

10:36 AM

My thinking was, if it does have a use, maybe someone here has heard of it.

But I guess not.

11:06 AM

Anyone good in geometry here? Can someone please help provide a rigorous solution to this question math.stackexchange.com/questions/1762456/…

Tobias Kildetoft

11:36 AM

If $A$ is a square matrix then clearly $AA^T$ and $A^TA$ have the same characteristic polynomial since these are conjugate. But what if $A$ is not square? A few examples suggest that they should have the same characteristic polynomial up to a power of $x$, but I have no idea if this is true in general and if so how one would prove it.

Hi guys! Does anyone know what are the physical applications of the Tree Matrix Theorem in Graph Theory?

@TobiasKildetoft have you heard of cayley graph before

Tobias Kildetoft

@Adeek sure

what is Draw the Cayley graph associated with (D10, {a, b})?

Tobias Kildetoft

@Adeek which elements are $a$ and $b$?

Wait, are you asking what it means to draw it?

11:50 AM

they are involution I guess

satisfying $a^2 = b^2 = 1$

Tobias Kildetoft

you can't just guess, you have to know if you want to draw the graph using them

yes that is right

those are the elements

Tobias Kildetoft

still not sure what you are asking

I am not entirely sure either

1 sec

@skillpatrol hmm, suspeciously silent depart

11:57 AM

@Adeek How was your presentation?

pretty good

@TobiasKildetoft so it is said draw cayley graph of $D_10,{a,b}$ where $(ab)^5 = e$ and $a^2 = b^2 = e$

do you have an idea what is that ?

Tobias Kildetoft

@Adeek just look up Cayley graph

ok

@JessyCat for cats i just need fish

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

but this is not just any cat; it's the cat from wonderland :P

12:13 PM

Hey @TobiasKildetoft

Can I see who downvoted my post ?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

nope

the Cayley graph of a group G with a relatively generating set S

Vertices elements of the group, and there is an edge between i and j if i = js for some $s \in S$

Why so?

Tobias Kildetoft

@Adeek More precisely an arrow, but yes

though I suppose one can just consider the undirected version

12:15 PM

now I am given D10 is generated by element of set S x, y satisfying a
x^2 = e
y^2 = e and
(xy)^5 = e. Draw the Cayley graph associated with (D10, {x, y}).

I don't even know how is the geometry of D_10

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

voter anonymity @Mambo is the network policy

Tobias Kildetoft

@Adeek What geometry? You just put the elements as dots and draw lines between them given by the above recipe

so $D_10$ has 5 elements right ?

Tobias Kildetoft

@Adeek No, 10.

isn't it $D_2n$

$D_{2n}$

Tobias Kildetoft

12:17 PM

@Adeek sure, but $n$ is not the order, it is the degree (which is unimportant here)

I see

so in this case it has 10 elements

so point $x \in D_{10}$ is connected to point $y$ by an arrow if it satisfies the relation above ?

and so we draw arrow between them ?

Tobias Kildetoft

right

but how can I find that the elements in $D_{10}$ are abstract

like how can I find that relation is satisfied for arbitrarily element $x,y \in D_{10}$

Tobias Kildetoft

@Adeek Well, presumably you are already familiar with the group. Otherwise, you have been given the generators and relations

the elements are all of the form $xyxy...$ or $yxyx...$ of some length at most $5$ (and note that $xyxyx = yxyxy$)

Any functional analysis people here?

Tobias Kildetoft

12:22 PM

and in fact all the elements but that longest one can be written in that form in a unique way

I see

Tobias Kildetoft

@Adeek Anyway, I need to go now

alright cya thanks a lot

12:33 PM

@Agawa001 indeed

12:49 PM

Are there examples where fractional calculus is used I other parts if mathematics except I guess differential equations?

1:39 PM

Hello any moderator here

1:56 PM

you have to ping one

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