Mathematics - 2015-12-01 (page 2 of 4)

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11:00 AM

Jasper can detect even the infinitesimal spelling mistakes.

4

user174558

I will start studying math on New Year's Day. I will begin with A first course in mathematical logic and set theory by Michael O'Leary.

user174558

I hope to complete one book per month next year. But I really hope I can start next year and not wait till next next year.

user174558

By the way, I did not star the above line by Balarka.

I'm mostly just an idiot

Is this still commutative algebra, or is it homological algebra?

in Commutative Algebra, 49 mins ago, by Robert Cardona

Daily Problem: Let $0 \to M' \to M \to M'' \to 0$ be an exact sequence. Show that if $M'$ and $M''$ are finitely generated, then so is $M$. Find counterexamples where you allow one or both of $M'$, $M''$ to not be finitely generated.

yes?

11:04 AM

@evinda i cant get the wind function working well in my matlab :S

I'm mostly just an idiot

@anon Yes it's commutative algebra?

user174558

@Agawa001 You need to open the windows to get the wind.

@I'mmostlyjustanidiot homological algebra and commutative algebra seem to have nontrivial intersection with exact sequences of modules. I'm not confident to make pronouncements though.

I'm mostly just an idiot

@anon Okay, thank you, can I ask you for advice on my attempt to solve this, assuming I make progress? (In 20 minutes or something)

@JasperLoy im not in the mood of morbid jokes

11:08 AM

@I'mmostlyjustanidiot it's 5am here, so if you do have an attempt it would have to be simple and clean for me to want to read it. I'm only up cuz I can't sleep.

I'm mostly just an idiot

@anon Oh okay sure, I'll probably wait for Robert to return then, I don't think I can make it clean on a first attempt :D.

Do you know the definition of an exact sequence is? That should be all that you need to do this problem.

I'm mostly just an idiot

@RobertCardona I do now(I'm in the CA room btw)

11:48 AM

Almost all the problems in the algebraic topology tag are either too easy or too hard for me to answer. :(

Not that I follow the tag very regularly...

user174558

If it is too hard, add some water to soften it.

user174558

I just watched Wrong Turn 1,2,3,4,5,6. All 6 movies were full of violence.

Watch Stalker.

user174558

Is it a movie about a stalker?

No.

I'm mostly just an idiot

11:58 AM

@JasperLoy Gross. Do some mathematics instead?

user174558

I was listening to Jackie Evancho on youtube. 10 year old girl who sings well.

user174558

@I'mmostlyjustanidiot Did you watch them as well?

@I'mmostlyjustanidiot Mathematics can be violent.

I'm mostly just an idiot

@JasperLoy I don't watch horror movies.

@BalarkaSen It's history especially!

user174558

@I'mmostlyjustanidiot I see. Are you a girl?

11:59 AM

E.g., consider the Casson handle.

I'm mostly just an idiot

@JasperLoy No, does not liking horror movies make me feminine?

user174558

@I'mmostlyjustanidiot I just speak my mind freely. No.

I'm mostly just an idiot

@BalarkaSen Violent as in sporadic?

Violent as in lots of blood, guts, etc.

user174558

I wonder if Agawa knows the meaning of morbid.

12:00 PM

So is the Alexander horned ball.

I'm mostly just an idiot

Why are these violent @BalarkaSen? I looked at pictures just now.

user174558

My mum is back, time for dinner, bye.

It was a joke :P

I'm mostly just an idiot

@BalarkaSen I don't get it :(

Pathological topological spaces are in general considered gory.

do you know the jordan curve theorem?

I'm mostly just an idiot

12:07 PM

@BalarkaSen No I am afraid not, where should I start reading to learn it?(Note that I am on a gap year, so I have a year to explore math freely) I'll come back and laugh at your joke :D)

I forget the proof of Jordan curve theorem using fundamental groups.

@Frank Me too. I only remember the homology proof, which is stronger.

@I'mmostlyjustanidiot You need to know some topology to understand the statement.

Homology proof is about duality theorem.

The proof is more complicated, which involves algebraic topology.

You mean the Alexander duality? You don't need that, not really.

I should really review the $\pi_1$ proof of JCT.

Trying to work it out alone may be interesting.

12:15 PM

Hmm, good point.

Conceptually, since your homology proof will involve things like $H_1$ to describe one dimensional things. Using $\pi_1$ is just another language, but expressing the same idea underlying, I suppose.

I doubt. $\pi_1$ of disconnected spaces make no sense, while $H_1$ is defined for a wide variety of spaces.

Just like MV sequence vs. Seifert-van Kampen

So it's easier to use $H_1$ to detect connected components.

Yeah, of course I agree with you that there are similarities ($H_1$ is abelianization of $\pi_1$ for path-connected spaces!) and the tools to compute the two are about the same.

But $H_1$ is a lot neater, easy-to-compute thing with natural non-crazy higher generalizations.

12:40 PM

hello, i'm looking for an elementary way to see that the ideal (X^2+Y^2-1, Y^2+Z^2) in C[X,Y,Z] is a prime ideal, any hints?

Overly Excessive

1:07 PM

Can someone explain, $\frac{1}{4}(\frac{1}{4})^2 \ne \frac{1}{4}^4$ ?

@OverlyExcessive does \ne mean not equal to?

Overly Excessive

Yes

I don't understand why it is not equal

To me, $\frac{1}{4}(\frac{1}{4}) = \frac{1}{4}^2$

@OverlyExcessive how many times have you multiplied 1/4 with itself on the left hand side?

@OverlyExcessive twice in (1/4)^2 and in front of it there is still one 1/4, so three times?

Overly Excessive

@happyEddie Well I understand that, but look, if $\frac{1}{4}(\frac{1}{4}) = \frac{1}{4}^2$ then we should be able to rewrite the first expression to $(\frac{1}{4}^2)^2$ and then it follows that $(n^m)^n = m^{mn}$ right

@OverlyExcessive i can't see why you can write the left hand side as ((1/4)^2)^2

Overly Excessive

1:14 PM

@happyEddie Because $n*n = n^2$

@OverlyExcessive that is correct

Overly Excessive

@happyEddie And what we have here is $n(n)^2$ So shouldn't that be equal to $(n^2)^2$ ?

@OverlyExcessive but the left hand side is not (1/4)^2 * (1/4)^2

@OverlyExcessive no, there is your misunderstanding

@OverlyExcessive n(n)^2 does not equal (n*n)^2

Overly Excessive

@happyEddie You mean because $\frac{1}{4}^2 = 1/16 \ne 1/4$ ?

Ah I see now

@OverlyExcessive in the first expression you square only n, in the second you square (n*n)

Overly Excessive

1:16 PM

@happyEddie I think I see it now, thanks Ed

I'm mostly just an idiot

Dumbish question: How do I verify an ideal of a ring $A$ is an $A$-module? Show that the 4 properties are satisfied?

@OverlyExcessive no problem

Overly Excessive

Intuitively I understood it but I couldn't get my math to add up, but now I see where I made the error :) .

Yes, any ideal has a natural $A$-action by multiplication.

You can easily show that this gives you a module structure.

I'm mostly just an idiot

I thought so.

Let $A$ be a local noetherian ring with maximal ideal $m$, then $m$ is an $A$-module and the action of $A$ on $m/m^2$ factors through $k=A/m$. What does this factors through $k$ mean?

1:20 PM

Presumably that the $A$-action on $m/m^2$ descends to a $A/m$-action (act by cosets).

user174558

1:30 PM

Hi @robjohn, I am now dodgerblue.

I'm mostly just an idiot

Hi @JasperLoy, how are you?

user174558

@I'mmostlyjustanidiot Same, not too good.

I'm mostly just an idiot

@JasperLoy That's not good. What are you doing right now?

Overly Excessive

Is it a rule that if $8 = 2^3$ then $2^6 = 8^2$ ?

user174558

@I'mmostlyjustanidiot Talking to you, lol.

I'm mostly just an idiot

1:32 PM

@JasperLoy Are you not multi-tasking?

user174558

@I'mmostlyjustanidiot Nope.

Overly Excessive

I mean generally for any integer where $b^n = a$

user174558

@OverlyExcessive You just need to know that $(a^m)^n=a^{mn}$.

I'm mostly just an idiot

I think he was told that one a few times now.

Overly Excessive

@I'mmostlyjustanidiot Huh :P ? What are you saying

user174558

1:34 PM

As for why it is true, you can see that easily by counting the number of copies of $a$ you have on each side.

I'm mostly just an idiot

@OverlyExcessive I mean I said that rule to you twice :P.

user174558

You see, once you really understand this rule, you don't even need to remember it or ask this question.

user174558

Like I said, practice is useless without understanding.

I'm mostly just an idiot

@OverlyExcessive $8=2^3$, $(8)^2=(2^3)^2=2^{2\times 3}=2^6$

user174558

Once you understand, end of story. QED.

Overly Excessive

1:36 PM

Ahhh.. I see

user174558

Read everything I just said with understanding, end of story.

Overly Excessive

@I'mmostlyjustanidiot I hadn't thought of applying the rule like that, but I see now

I'm mostly just an idiot

@OverlyExcessive That's good :).

Now if only I could understand the answer Balarka just gave me!

user174558

It is not good enough to remember the rule.

Overly Excessive

@JasperLoy It's taking me some time to internalize the concepts, I'm always unsure whether I can generalize them or not.

user174558

1:38 PM

@OverlyExcessive Take your time, it's OK. Many math teachers know shit about these things too.

@JasperLoy: =(

user174558

I have come across too many incompetent math teachers in my country.

Overly Excessive

@JasperLoy I see what you are saying though, I am trying to understand the concepts behind these things instead of just rote memorization of techniques and rules.

user174558

One said that the empty set is the same as the set containing the empty set, enough said.

user174558

@OverlyExcessive Well said.

I'm mostly just an idiot

1:40 PM

Could you expand on this at all? I need to understand it to understand another statement:

"A local noetherian ring is regular if and only if it's Krull dimension is equal to the dimension of the vector space $m/m^2$"

It is saying that the $A$-action on $m/m^2$ factors to a vectorspace?

the problem is to find the irreducible components of V(X^2+Y^2-1,X^2-Z^2-1) in C^3, first i thought the set is irreducible but now i notice that Y-iZ or Y+iZ is not in the ideal (X^2+Y^2-1,X^2-Z^2-1) but their product Y^2+Z^2 is so the ideal is not prime, hence the set is not irreducible, how would you go on searching for the irreducible components?

Overly Excessive

Have they ever had to redefine something in math? Is that even possible?

Tobias Kildetoft

@I'mmostlyjustanidiot Well, yes, by definition, $m$ acts as $0$ on that quotient, so the action of $A$ factors through to $A/m$.

@I'mmostlyjustanidiot you view m/m^2 as a module over the field A/m, i.e. a vector space

@I'mmostlyjustanidiot I mean, if $A$ acts on something, then you can get a natural action of A/m if elements of m act trivially.

The action is $A/\mathfrak{m} \times M \to M$ given by $(a + \mathfrak{m}) \cdot p = ap$.

1:49 PM

btw, can you see this latex code in the chat in rendered form in some way? i find it very hard to read in this form

See the message on the star panel, starred by 10 chatters, labelled "$\LaTeX$ in chat"

@BalarkaSen thanks

2:02 PM

@BalarkaSen We don't need auto corrects/grammar checks .. more Jasper is what we need! :P

1 hour later…

3:28 PM

A subsequence being terms $k_n\le k_{n+1}$ is a "wrong definition" right? It should be $k_n<k_{n+1}$ (it does actually matter, define $k_n:=1$ then every sequence has a convergent subsequence) so it /should/ be the latter and there's no case for the former

@DanielFischer Hi!!! I have thought about what you told me yesterday... I think that it is right as follows, since we pick an arbitrary element in $\ell^2(\mathbb{N})$ and justify why $Y$ is dense in $\ell^2(\mathbb{N})$, which means that ||y-x|| ->0 for any y in Y. Do you agree?

Let $x=(x_1, x_2, \dots, x_n, \dots) \in \ell^2(\mathbb{N})$.

Then $\sum_{k=1}^{\infty} x_k^2<+\infty$.

We fix a $ n \in \mathbb{N}$ and have the following:

$\sum_{k=n+1}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^n x_k^2$

Good morning

3:49 PM

Hi @Overflowh

@evinda hey hello.

How are you? @Overflowh

@evinda Demoralized and discouraged. And you?

Why? @Overflowh

Heya all! :)

4:05 PM

Because I have an exam in 8 days and I though I was understanding "things" but turns out I wasn't since I can't solve the damn exercises :/ (and also because the practice exercise I was able to solve seem to be one tenth of the difficulty of the exam exercises...) @evinda

Hello @PerplexedGuest

4:17 PM

@BalarkaSen: dumb question, $\Lambda^1 T^*M \neq T^*M$, right? what exactly does this lambda do again? ._.

@Overflowh In which subject?

$\Lambda^1 T^*M$ means $1$-forms on $M$?

ah gee

I keep getting confused at notation

ofc

@evinda Linear Algebra, not even that complex algebra actually

Things like these: mathb.in/47589. I don't have have idea where to start. I'm trying to solve it anyway, but it seems my solution are different from the one provided (that aren't granted to be actually accurate either).

4:38 PM

hi all

@Huy The first exterior power of $V$ is $V$.

Remember that the exterior power is basically "$n$-tuples, but we skew-symmetrize"

Well, if you're working with 1-tuples, there's not two things to skew-smmetrize.

@anon: No idea.

Seems like a neat construction.

Hi

5:07 PM

Okay! so can I get some help in Information theory here?

@Overflowh.: hey! can you help?

@coderredoc I'm the last of the newbies in here, I'm afraid I have no idea how to help you, sorry.

hey it's okay..no big deal :)

Mike ?

@MikeMiller

6:13 PM

@MikeMiller: but why would anyone bother writing it down then?

To put it in context with the other exterior powers?

@Huy if a set $A$ contained in a metric space is totally bounded then will it's intersection with an open ball also be totally bounded?

6:43 PM

@Paradox101 isn't any subset of a totally bounded set totally bounded?

@happyEddie that's what I thought but my instructor said to check it again so now I'm a little confused

If the totally bounded set intersects an open ball that intersection will be contained in the set $A$ and since $A$ is totally bounded the intersection should be too

@Paradox101: it's always good to check again just to be sure

doesn't mean it's not true

how would you prove it?

@Huy to prove it should I consider a ball with center say $x$ and then show what the intersection would be if $x$ was an interior and exterior point in the metric space?

6:59 PM

I am not sure what you mean by that, I only know of the definition of totally bounded with epsilon balls

so I'd just apply the definition

@Huy I was just wondering whether the open ball in question would be completely outside or inside $A$. If we simply say that the intersection would be covered by a collection of epsilon balls and hence is totally bounded would that simply make up the proof?

the open ball? which one do you mean?

can you write down precisely what it means to be totally bounded?

@Huy the open ball intersecting with $A$

a set $A$ will be totally bounded if for any epsilon there exists a finite epsilon-net for $A$

ok

i.e. that the set will be covered by a collection of open balls with radius epsilon

7:06 PM

ok

now again about subsets

can you show that a subset of $A$ is also totally bounded?

btw, your last statement is missing "finite"

@Huy doesn't that simply follow from the fact that a subset is contained in a set and if that set is covered by a collection of balls the subset too will be covered?

yes, that's true

finite, please

:D

oh sorry I forgot to add that

finite collection of balls :D

any more questions ?

Tobias Kildetoft

What is the difference between bounded and totally bounded?

7:08 PM

so will that simply be the proof?

isn't it a trivial one?

@TobiasKildetoft: are you asking or are you asking Paradox101?

Tobias Kildetoft

@Huy I am asking.

@TobiasKildetoft: unit balls are bounded but in infinite-dimensional spaces not totally

Tobias Kildetoft

@Huy Ahh, of course.

@Huy is an empty set totally bounded?

7:11 PM

@Paradox101: what do you think? :P

(Bounded => Totally bounded) => locally compact

@Huy it is? :p

i mean nothing can be covered with anything?

we can always find any epsilon whose finite collection of open balls will cover nothing

wut, I had 2 upvotes on this answer, now I have 7.

is the empty set in the empty metric space totally bounded?

who the hell would care about such a thing :P

7:17 PM

no?@happyEddie

i don't know, i think it depends on the definition

if we must have a collection of open balls covering the empty set, then it is not, because there is no open balls in the empty space

yes exactly, but how would it be wrong if we took a different definition?

Tobias Kildetoft

@happyEddie The empty collection of open balls is still a collection of open balls

if for any epsilon we must have a finite cover with radii of its elements at most epsilon, then we can take {emptyset} as the cover of the empty set

@TobiasKildetoft a-ha, you are correct! i missed that

so the empty set will be totally bounded in any case

@TobiasKildetoft how is an empty collection of open balls still a collection of open balls? I mean how will an empty collection cover and empty set?

7:26 PM

this is getting confusing

Tobias Kildetoft

@Paradox101 vacuously

lol

we had this theorem today: let $G$ be a Lie group. Then, if G is solvable, g is solvable and if g is solvable and G is connected G is solvable.

it works better when not written

ahaha

Tobias Kildetoft

@Huy So many g's when you say it

7:27 PM

@TobiasKildetoft oh ok. This was interesting

@Paradox101: what course is that btw? analysis 1?

@MikeMiller $[0, 1]^2$ and the comb space inside it is a counterexample to this, isn't it?

trying to wrap my mind around the Poincare conjecture. Can anyone help? math.stackexchange.com/questions/1113810/…

no, that's not what Poincaré conjecture states

By comb space, I mean $[0, 1]$ attached to each pt in $\{0\} \bigcup_n \{1/n\}$.

7:32 PM

@DanielFischer I asked yesterday something about uniform convergence of functional series $\sum_{n=1}^{\infty} \frac{\sqrt{1+2^nx}}{n!}$ on $[0,\infty)$, and you said that this series is not uniform convergent, because that part with $\sqrt{\cdot}$ is unbounded. But, now, I am little bit confused, because we have that $n!$, so, so...

Specifically how do I do this: "A prism is the boundary of a convex subset of R3
R3
so just pick a point on the interior and project along a line from that point onto a sphere which is large enough to enclose the prism. This will define a homeomorphism from the prism to the sphere."

Inflate the prism. You get a sphere.

:(

How do I inflate?

user174558

@BalarkaSen Do you know why Sayan deleted his account?

By pumping air inside it.

7:34 PM

seriously tho :(

@JasperLoy Nope.

@Huy it's analysis 2

user174558

@BalarkaSen Might be better for him that way, then he can really finish Apostol. He claims he did all the exercises but I think he is lying.

Or maybe he got demotivated because we were ignoring him, and gave up studying math. I was planning on getting his e-mail from Soham.

user174558

Then he is not really interested anyway.

7:38 PM

^

@JasperLoy My thought exactly.

user174558

@happyEddie Do you know him by the way?

Usually Mike and Ted's ignore pushed me to study math harder (not complaining!), but maybe there are people who get demotivated instead of more motivated. Or maybe you're right.

@JasperLoy No, I actually have no idea about whom you are talking. Anyway I can't see how one's interest in maths can depend on how other people relate to one.

user174558

@BalarkaSen I see. People's ignore or not has no effect on my math.

7:42 PM

@happyEddie: frustration can reduce motivation, but I think for example some competition is usually very healthy

@Huy I don't know. For me, studying and competing don't mix. I study only to learn. Competing or showing off with my knowledge are the last thing in my mind when I study.

no, I don't mean showing off. I mean thinking "wow this guy knows so much about maths I need to study a lot harder so I can know as much maths as he does one day"

2

My opinion is that if you study maths in order to impress someone else you are studying it for wrong reasons.

^

by ^ I meant to ^ Huy's message

Tobias Kildetoft

@happyEddie If you want to stay in academia you will need to change your mind on that at some point :)

7:51 PM

@happyEddie nobody would learn math just to impress someone. that's silly. but being recognized as "knows something" adds to your motivation.

@TobiasKildetoft Good point. That is one reason I'm only a hobbyist in maths.

@Cortizol But $n!$ doesn't depend on $x$. For every $n$ we have $\lim\limits_{x\to +\infty} \dfrac{\sqrt{1 + 2^n x}}{n!} = +\infty$.

@BalarkaSen OK, I can't deny that.

@evinda "$\lVert y-x\rVert \to 0$ for any $y\in Y$" doesn't make sense. What you want to say is that for every $x$ there is a sequence $(y_n)$ in $Y$ such that $\lVert y_n - x\rVert \to 0$.

over-recognition would inflate your head (speaking from personal experience) and let you slide back to laziness, however. I only mean that it helps if you know you might not be as bad as math as you thought.

replace "math" by anything you're interested in doing, really.

7:55 PM

yea it works for everything

Yes, it's good to know where you stand. However, why do you need other people to tell you that?

nobody said that?

@BalarkaSen talked about (over-)recognition

community recognizes whether what you did is dumb and of no use or whether they are of a tiny importance. you can't evaluate yourself (well, I can't) - that's a form of social exam.

It's good to have a correct picture in your head about what you can and can't do. About what you are good at and where you need to improve. However you don't necessarily need other people in this.

Why can't you evaluate yourself?

7:59 PM

instincts. you stick to what you think is good, but it might not be what mathematical community thinks is good.

Tobias Kildetoft

@happyEddie The many people who upload to vixra show why you can't evaluate yourself whether you are good at math

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