Mathematics - 2015-11-04 (page 2 of 3)

Alizter

17:06

hey @BalarkaSen

Balarka Sen

17:21

hi @Alizter. I'm sick :(

Alizter

:(

Here is a problem for you then

Should be simple

Balarka Sen

so, what's up?

ok. "simple" depends on what the problem is.

Alizter

Let P be an integral polynomial, prove that there exists an integral polynomial Q such that $P\circ Q$ is reducible for all P.

Balarka Sen

very badly stated.

Alizter

How should I state it?

Balarka Sen

17:23

prove that there is a polynomial Q with integral coefficients such that P \circ Q is reducible for all polynomial P with integral coefficients.

not the kind of problem i care about. if i ponder and don't find something immediately, i am going to stop thinking about it.

sounds like an olympiad problem to me.

Alizter

I think it is from an algebra book

Balarka Sen

doesn't sound very algebra problem to me.

Alizter

I have a proof

Balarka Sen

1 sec.

meh, dunno. don't care. tell me your proof.

Alizter

Say that $P\circ Q$ was irreducible, then by Eisensteins criterion, there exists no prime $p$ such that $p$ divides the leading coefficient of $P\circ Q$. Let $\deg P = n$ and the leading coefficients of $P$ and $Q$ be $a_P$ and $a_Q$ respectively. Then the leading coefficient of $P\circ Q$ must be $a_Pa_Q^n$ which means $a_P=a_Q=1$ if it is divided by no prime $p$, which is not true as all $P$ where considered and therefore we have a contradiction.

Balarka Sen

17:29

i doubt your proof is correct. you have to prove that there exists such a Q, right?

where have you proved that?

Alizter

By showing that if such a Q exists then it has to be reducible

hmm no proof of Q existing thouhg you are right

Balarka Sen

told ya.

Galois in the Field

How do I find the fraction field of $\Bbb Z[i]$?

Balarka Sen

so no proof.

@GaloisintheField $\Bbb Q(i)$

Alizter

@GaloisintheField How did you find the fraction field of $Z$?

Galois in the Field

17:30

Yes, but how do I find that?

Balarka Sen

just consider fractions of Gaussian integers

then do algebra

i'm having dinner now, bubye

Galois in the Field

@Alizter I formed equivalence classes $\Bbb Z\times \Bbb Z/ -\{0\}$

Where $(a,b)\sim (c,d) \iff ad-bc =0$

$(a,b)=a/b$

Balarka Sen

@GaloisintheField you can derive an explicit isom here too

Galois in the Field

Explicit injective homomorphism?

Balarka Sen

just try it out

Galois in the Field

17:33

$\psi(a)=(a,1)$

Balarka Sen

i mean isomorphism.

Galois in the Field

Between what?

Balarka Sen

nono, i mean isom between frac Z[i] and Q(i)

Galois in the Field

Oh ok

I can't immediately see why

Oh ok

Yep

Alizter

You got it now

Balarka Sen

17:39

@Alizter Have you figured my problem out?

Galois in the Field

Yeah $\psi:\Bbb Z[i]\times \Bbb Z[i] \to \Bbb Q[i]$ is surjective since $\frac ab + \frac cd i=\psi((a,b)+(ci,d)=(ad+bci,bd))$ hence the identity map is an isomorphism

Oh I did something illegal, but oh well, I'll move on

Balarka Sen

@GaloisintheField That map is not a ring homomorphism.

Galois in the Field

Yeah I screwed up something chronically

I'll do it tomorrow when I'm not tired

It's 5am in NSW australia

Balarka Sen

Get some rest.

Galois in the Field

Good luck pals

Balarka Sen

18:00

Morning @MikeMiller. I was too sleepy yesterday : I agree that I should have been careful with my statements especially since a topologist had been saying I was wrong. Thanks, I will be more careful from now on.

Mike Miller

:p

Agawa001

@BalarkaSen i agree, especially with the term (eliminating variables)

Balarka Sen

I have not been able to do much math, but I have thought a little bit about the duality between transversal intesection of submanifolds and cup product. I don't think I have a proof, but just an intuitive idea. Do you think you would like to listen, @MikeMiller?

Mike Miller

Probably not, sorry. Good luck with it.

Balarka Sen

Ah, no problem.

I believe I "understand" why they should be Poincare dual.

Mike Miller

18:08

Tell me another time when you think you've got it and when I have time.

Balarka Sen

Alright, thanks very much. I am trying to thrash out the details.

Alizter

18:27

@Khallil Have you chosen a topic for your second year essay?

@BalarkaSen Which one?

Balarka Sen

index 2 subgrp normal

Alizter

oh yes, I haven't had time to work on it

I shall work on it now

Balarka Sen

Sure :)

I will give you harder and more interesting problems afterwards.

Paradox 101

Can anyone explain whether taking different tags changes riemann sums?

Alizter

@Paradox101 What are tags?

Paradox 101

18:36

@Alizter I'm talking about the tags which give rise to tagged partitions through which we define the riemann sum

Chris's sis the artist

PAM-PAM-PAM

Guess what?

@Agawa001

Agawa001

@Chris'ssistheartist another nice achievement !

Alizter

@Paradox101 interesting question.

Chris's sis the artist

@Agawa001 I prepared the generalization for a magazine.

@robjohn see the deleted thing above. I prepared that generalization for a magazine.

Agawa001

cool, i ever wished to work for a magazine publishing

Chris's sis the artist

18:40

Actually, I have to finish another problem to that magazine too, that is because I want to related another very nice problem to one that I previously proposed.

@Agawa001 Maybe I set up my journal and I invite you to work there (sometime in the furture). :-)

Paradox 101

@Alizter i know that taking different partitions doesn't change it but I don't know about tags. I still don't completely understand the concept of tags

Chris's sis the artist

The only bad, very bad thing is that few people think like me :-( (I don't mean the smartness, but the dedication to this stuff)

Agawa001

@Chris'ssistheartist wont be satisfied of any post apart ceo

Alizter

@Paradox101 From what I quickly read, I understood tags to be an arbitaryly size interval partition

Chris's sis the artist

@Agawa001 OK OK, good to know that! :D

Alizter

18:43

I guess as long as they all $\to 0$ it shouldn't change much about the riemann sum

But I wouldn't know how to prove that

Paradox 101

@Alizter tags alone were any points on a particular sub-interval

Balarka Sen

I just finished watching The Unbreakable. Great movie.

Paradox 101

whereas a tagged partition is an ordered pair of sub-intervals and their corresponding tags

Agawa001

@Chris'ssistheartist next month i ll pass thru a job contest for accountants

Chris's sis the artist

@Agawa001 Really?

Agawa001

18:47

@Chris'ssistheartist the problem is, i dont want to be accountant

Chris's sis the artist

@Agawa001 Well, then try something else

Agawa001

@Chris'ssistheartist i have no option

morphic

@BalarkaSen Which year is the movie from?

Agawa001

accountant seems to be a harsh way for gradual suicide

Balarka Sen

I don't know.

Agawa001

18:53

@Chris'ssistheartist i have 26 y o now, by the next year, job opportunities will reduce by more than half, i must take that risk, i ll no longer wait for the (golden suitable luxurious job)

Balarka Sen

hi @TedShifrin

Ted Shifrin

hi @Balarka @Agawa

Chris's sis the artist

@Agawa001 I know your point.

Ted Shifrin

heya mr eyeglasses

Agawa001

hi @TedShifrin

Ted Shifrin

18:54

I don't blame you, @Agawa: I could not stand to be an accountant, either.

Agawa001

@TedShifrin what a tragic end. :'(

Balarka Sen

@TedShifrin I believe cantor function restricted to the cantor set is the desired surjective map.

Ted Shifrin

Depends what you mean by that, @Balarka. But probably yes.

It's quite easy to be explicit.

Balarka Sen

Admittedly, I didn't find it all by myself. Probably wouldn't even have thought of it if Khallil had not mentioned it.

Ted Shifrin

In fact, when you were trying to show off, you essentially were. ... Then do you finish with the space-filling curve?

Mr. 2-adic show-off? Ahem.

Balarka Sen

18:56

oh? what has the 2-adics got to do with it? ohh.

Yes, over with the space-filling curve.

Ted Shifrin

rolls an extra half eye and all 8 this time

Balarka Sen

@TedShifrin You need to see a doctor.

Sprouting eyes after retirement is not good.

Ted Shifrin

OK, then, back to Chapter 5 or diff geo.

morphic

ಠ_ಠ

Ted Shifrin

mr eyeglasses lent me his glasses, so it's ok.

Alessandro

18:58

good evening @Ted

Ted Shifrin

hi @Alessandro

Balarka Sen

@TedShifrin You need an extra-cool eyeglass to cover your 8 eyes all at once.

Ted Shifrin

You worry about learning math, @Balarka, and I'll cope with my eyes.

Balarka Sen

Well, I'm sick!

Ted Shifrin

But that's a permanent state with you.

Balarka Sen

19:00

Not really.

Ted Shifrin

We should check the chat records.

Balarka Sen

The headache just came over today evening.

Alessandro

I'm trying to understand what the dual space of a vector space is @Ted but I find it very confusing (possibly because we haven't talked about linear functions yet...)

Paradox 101

Can anyone explain the highlighted part of this answer?

Ted Shifrin

But you know what linear functions are, @Alessandro. Have you learned dot/scalar products yet?

Alizter

19:02

Ah @BalarkaSen I did it

Ted Shifrin

What don't you understand, @Paradox?

Alessandro

well, I read what a linear function is, but we didn't talk about it in the course yet (dual spaces came up in a more advanced non-mandatory lecture)

Balarka Sen

@Alizter OK?

Alizter

Let $H<G$ and $|G:H|=2$ then $g\in G$ we have $gH$ equal to either $Hg$ or $H$. As it is not equal to $H$ then it must be equal to $Hg$.

I though too long about it but I was trying to be careful

Ted Shifrin

@Alessandro: A linear function $f\colon\Bbb R^n\to\Bbb R$ is of the form $f(x)=a\cdot x$ for some vector $a\in\Bbb R^n$.

Alessandro

19:04

Actually I think I understood the concept of a dual space and the construction of the dual basis, what I'm missing is its significance

Paradox 101

@TedShifrin I don't understand why $M_1=1$. I know that $M_1=sup f$ but it's the sup over interval $1$ and if interval 1 is as mentioned in the answer then the value of $f$ there is 0 so $M_1$ ought to be zero

Alizter

@Alessandro Row and coloumn vectors

Ted Shifrin

@Paradox: You consider $f(x)$ for all $x$ in that interval. Since $f(0)=1$ and that's the maximum value, you get $M_1=1$.

@Alessandro: It turns out that studying the vector space of all linear functions on a vector space is a very interesting thing (especially in infinite dimensions). And multilinear functions (tensors, determinants, then differential forms) are super-important in math and physics.

Balarka Sen

My internet is horrific. Yes, @Alizter, that is correct.

Alessandro

What is really confusing to me is how the professor got a linear function "for free" (without having to talk about basis or anything) from $V$ to $V^{**}$

Alizter

19:06

@TedShifrin Especially in physics, pretty much everything is a tensor in physics.

@BalarkaSen Another problem?

Balarka Sen

@Alizter How about - classify all groups of order $4$ upto isomorphism?

Paradox 101

@TedShifrin ok I get that part, but why are we also including 0 in interval 1? I mean, it's a discontinuous function so shouldn't we take separate intervals for the two different parts of $f$?

Alizter

@BalarkaSen Klein 4, and Cyclic 4

Balarka Sen

Proof?

Alias User

Hello all! Why is it that a Poisson distribution requires p to be small to approximate a binomial distribution?

user159870

19:08

Hello @TedShifrin ! I have a question to you...

Ted Shifrin

Well, but they needn't be the same, @Alessandro, but yes: There's a symmetry in whether a dual vector eats a vector or a vector eats the dual vector. See $f(x)=a\cdot x$. This is just as well a linear function of $a$.

Huy

@Alizter: I recently ran into someone who has graduated in physics who didn't understand what tensors were. I have no idea how that happened.

Ted Shifrin

Riemann integrals are based on dividing the interval into closed subintervals, @Paradox.

The $i$th interval is $[x_{i-1},x_i]$, endpoints included.

Best thing to do is draw graphs, @Alias. If you have access to Mathematica, I have a notebook I wrote to show my students last fall when I taught probability.

user159870

@TedShifrin I am looking at the question: 1. Show that there is a sequence $T_1, T_2, T_3, \dots$ such that $T_1 > T_2 >
T_3 > \dots > 0$ and that $\gamma$ is $T_r$-periodic for all $r \geq 1$.

of this exercise: http://math.stackexchange.com/questions/1477087/how-to-show-that-such-a-sequence-exist

I have understood how we define the sequence. But could you explain how we show that $\gamma$ is $T_r$-periodic for all $r \geq 1$ ?

Ted Shifrin

What, @user159870?

user159870

19:11

@TedShifrin I wrote it :))

Alias User

Thanks @TedShifrin! Mathematically, though, I'm curious why the two distributions don't line up - the poisson is the limit of the binomial as time steps get indefinitely small, and I know there's the assumption that only one event can occur per timestep, but if the timesteps get indefinitely small, shouldn't that not matter? Couldn't p be anything?

Paradox 101

Ohhh ok. If we have another question similar to this one in which for the first part $x$ is between 0 and 1 (both inclusive) and for the second $x$ is greater than 2, with 2 inclusive, should we then assume different intervals for them both? @TedShifrin

Ted Shifrin

We already had a counterexample a week ago, @user159870, and André gave you the same one.

Alessandro

$V^{**}$ is the dual of the dual of $V$, so elements of $V^{**}$ are linear functions associating a linear function in $V^*$ to an element of the underlying field, right? @Ted

Ted Shifrin

Sure, @Alias, if you make the steps smaller and smaller. There's a uniformity question going on here.

user159870

19:12

@TedShifrin I mean if we suppose that it is also continuous.

Ted Shifrin

OK, @user159870, then it's true. You have lots of answers there. So what specifically do you not understand?

Alias User

@TedShifrin - what's the uniformity question?

Ted Shifrin

I don't understand, @Paradox.

I mean, @Alias, that step-size (if you like) and $p$ have to be correlated. You can't find a step-size that works for all $p$ unless they're all smaller than something.

Yes, @Alessandro. $V^{*} = (V^)^*$. Weird that this isn't showing up right.

Alias User

Thank you - that makes sense. So why would you use a Poisson distribution over a Normal distribution for approximating many trials of a binomial distribution?

Paradox 101

Balarka Sen

19:16

@Alizter Want a hint?

Paradox 101

If it's something like this then should we assume two separate intervals for both the cases? @TedShifrin

Alizter

@BalarkaSen no

Huy

and what if $x \in (1,2)$?

Balarka Sen

@Alizter OK. How're you approaching it?

Ted Shifrin

Good question, @Alias. It's really a question of the interpretation of the random variable. Poisson shows up with certain sorts of applications. I'm sure you can find this in your text or on line.

Alizter

19:16

Lagranges theorem

Balarka Sen

(my hint would have been another problem, actually).

@Alizter Good idea.

Alizter

$x^4=1$ forall x in G

Ted Shifrin

@Paradox: You need a function defined on the entire interval $[a,b]$. You can't have a gap.

Balarka Sen

@Alizter Not true.

Alizter

@BalarkaSen Yes true

That is lagranges theorem no?

Balarka Sen

19:17

? It is not.

The klein 4 group have no order 4 elements.

Ah.

Yes, that is true :P

Alizter

It is a direct Corollary of lagrange

Balarka Sen

Sure, sure. I misread $x^4 = 1$ above by "order of $x$ = $4$". Apologies.

Alizter

okok

Then I am looking at possible presentations

I could list them all

Balarka Sen

Go on.

No, I don't want brute force.

Paradox 101

@TedShifrin isn't this function similar to the original one? Both are discontinuous

Balarka Sen

19:19

If you do brute force, I'll ask you to classify groups of order $31^2$.

Alizter

or better yet 24

Balarka Sen

24 is not square of anything.

Alizter

But it has a lot of subgroups

Alessandro

I had to escape * with * too @Ted ok, so, we defined the function $h:V\to V^{**}$ that maps $v\in V$ to $g\in V^{**}$ where $g$ is the function that maps $f\in V^{*}$ to $f(v)\in\mathbb{K}$ and we want to show that this function is linear

Balarka Sen

@Alizter Yes, that is why you cannot classify it so easily. There are not many groups of order $31^2$. I'll tell you about it when you finish $4$.

user159870

19:21

@TedShifrin
For the first question.
We can show that there is such a sequence as follows:

If there is no such $T_0$ , by definition that means that for every $T_0>0$
there exists $0< T'<T_0$ such that $\gamma$ is $T'$-periodic.
We use this to define the sequence iteratively: to get $T_1$ , we take $T_0=1$
in the above, and we let $T_1$ be the corresponding $T'$ . Then, we "take $T_0 =T_1$ in the above, and we let $T_2$ be the new "corresponding T ′
. Etc.: given $T_n$ we define $T_{n+1}$ to be the T ′ corresponding to $T_0=T_n$.

Ted Shifrin

Read what I said, @Paradox. You only talk about integrals on an interval $[a,b]$, so the function must be defined on all of $[a,b]$.

heya @AlexG

Balarka Sen

@AlexanderGruber!

Alexander Gruber

bonjour

Ted Shifrin

Correct, @Alessandro, linear in $f$.

Balarka Sen

Whatcha upto?

Ted Shifrin

19:23

@user159870: It's just too crazy in here now. I can't deal with things so complicated and long.

Alexander Gruber

Miserably failing to C++ a thing

Paradox 101

@TedShifrin oh ok I get it. thank you :)

Alizter

@AlexanderGruber What is it

Alexander Gruber

@Alizter some kind of pointer bug. I'm bad with C++.

Balarka Sen

Here's a question that's bothering me (should be very simple) : If $A,B$ are abelian groups, $f : A \to B$ a homomorphism, then if $f \otimes \text{id}_{\Bbb k} : A \otimes_{\Bbb Z} \Bbb {k} \to B \otimes_{\Bbb Z} \Bbb k$ is an isomorphism for all fields $\Bbb k$, is it necessary that $f$ is an isomorphism?

user159870

19:27

@TedShifrin :(

Alexander Gruber

trying to populate this binary tree and it says it's adding nodes and then when i go back to print them it says they're empty

Balarka Sen

I think it is.

Alizter

@AlexanderGruber I though C++ was supposed to move away from pointers

Ted Shifrin

@user159870: The point is that $\gamma$ is $T_r$ periodic for every $r$. So you get a sequence $T_r\to 0$ with $\gamma$ periodic for all of them. What's wrong with that?

Alexander Gruber

@Alizter idk, i'm using them for speed

Alessandro

19:28

Ok, I think I understood that part @Ted. Now I have to understand how that came up in a combinatorics lecture...!

Alexander Gruber

it's the fundamental structure for this thing i'm going to have to do bajillion sized samples of so i'm trying to make it as fast as possible

Ted Shifrin

Ah, that I don't know, @Alessandro.

Alizter

@AlexanderGruber Whose data structure is the binary tree

Alessandro

me neither! @Ted

Alizter

Is it yours?

Alexander Gruber

19:29

Yeah

Alizter

Thats probably where the problem starts then

Ted Shifrin

But I hope you read more of Babai's notes, @Alessandro. I think they're amazing.

Alexander Gruber

undoubtedly

Alizter

Have you tried testing it for small ones

Alexander Gruber

yeah, that's where it's messing up.

Balarka Sen

19:29

@TedShifrin What's Babai's notes?

Alizter

Going through the program by hand

Alexander Gruber

yep

Alizter

I can't do much else if I don't see the code, but you don't have to post it it's up to you

Alexander Gruber

oh, well, if you're interested sure

Ted Shifrin

Linear Algebra applied to all sorts of fascinating combinatorics, @Balarka.

Alexander Gruber

19:30

pastebin.com/TmDqrhaa

output: pastebin.com/JMW8hDPD

Balarka Sen

oh, nice. What sort of combinatorics?

Alizter

@AlexanderGruber You may also want to consider using other peoples datastructures for trees

Ted Shifrin

google and look

Alizter

I am sure they are fast as well

Alias User

@TedShifrin - I apologize for so many questions, but I wasn't able to find a justification of when to use a Poisson distribution over a Normal distribution for binomial distribution approximation. Would you mind pointing me in the right direction?

Alessandro

19:31

I think I'll read more of it next week @Ted I'm too busy preparing for my exams right now

Alexander Gruber

@Alizter well, i'm also trying to actually learn what i'm doing. I shouldn't be this bad at C++

Alias User

It seems weird that a binomial distribution can approach both a Poisson and a Normal distribution as n goes to infinity if p is small.

Alessandro

time for dinner! I'll be back later

Ted Shifrin

@Alias: I'm not an expert in this stuff. But the text that I taught out of, Ross, certainly discusses it. Also, an excellent web-available reference, is a text by two MIT professors, Bertsekas and somebody.

Buon appetito, @Alessandro :0

Alias User

@TedShifrin - thank you!

Alexander Gruber

19:33

so see in the output, it says it adds the left and right nodes to (1,1), but then when I go back and try to print them it says they're both empty.

Balarka Sen

Oh, very cool, @TedShifrin.

I see that it does some chromatic combinatorics (Ramsey-type stuff) on 4.2.

Ted Shifrin

Also settles the Borsuk conjecture (more geometry) ... totally cool.

Balarka Sen

Let me google Borsuk conjecture, never heard of it.

Ted Shifrin

It's in the notes.

Alexander Gruber

oh here is the code with syntax highlighting @Alizter

Balarka Sen

19:39

Wow, ok. That sounds very nontrivial.

Ted Shifrin

Indeed. He has history in there.

Ghost

heh

It is not about your happiness - I figured the answer, but as with my students - you show an effort first, then I help — Ghost 2 mins ago

Alizter

@AlexanderGruber Could you try and even simpler tree maybe:

./\
/\

5 nodes

Balarka Sen

@TedShifrin Hmm, odd. I thought Erdos-Szekeres problem said that if I have some integer $n$, and there is a bunch of points in general position on the place with cardinality = $m$, then when is it true that there is a subset of $n$ points of that set which forms a convex polygon?

Maybe I am misremembering.

(By that, I mean, what is the bound $N$ on $m$ w.r.t $n$ such that it is true for all such $m > N$?)

I think Erdos and Szekeres proved that $N$ is finite.

Tien-Cheng Huang

Why my question was put on hold as too broad? math.stackexchange.com/q/1509214/275935 Could someone explain?

I want to put a bounty on it but now I can't

Ted Shifrin

19:50

Because you're asking for someone to type you out an hour lecture. You can find this in books. @Tien

Tien-Cheng Huang

@TedShifrin Which book? I can't find the procedure in D&F.

Balarka Sen

It's certainly in D&F.

They refer to the chapter on modules, I believe.

D&F does this more generally on modules over PID's.

Ted Shifrin

It's in there. Or look at Artin's Algebra. Or Jacobson's Algebra, volume I, I believe.

Alexander Gruber

@Alizter if I comment out two of the nodes before adding them to the vector, there's still one missing node

when I print

Alizter

Are you sure you are using the correct pointers when adding nodes or printing

Balarka Sen

19:52

Oh, turns out I was mixing up with the happy ending problem @TedShifrin.

Alexander Gruber

bpaste.net/show/5306fd5f39ae

Balarka Sen

Both are proved by Erdos and Szekeres, on the same paper :P

Alexander Gruber

@Alizter I think so. The control flow is right, at least. I put the couts in there for that reason

Ted Shifrin

I know none of this stuff, @Balarka.

Tien-Cheng Huang

I have D&F at hand now. Could someone tell me which page?

Ted Shifrin

19:54

There are so many editions, @Tien. That won't help.

Tien-Cheng Huang

most recent edition is 3rd edition

Balarka Sen

You certainly know more than me - I have only heard about these from my algebraic number theory professor who's interested in doing chromatic combinatorics. He's trying to apply dynamics in his work, he said, iirc.

Ted Shifrin

I only have the 2nd. Look in the chapter titled Modules over Principal Ideal Domains, @Tien.

Mike Miller

"they are Ricci flat and carry non-trivial parallel spinors which, so I am told, makes them relevant to high energy physics."

Balarka Sen

@MikeMiller If that was directed towards me - I never said that makes it relevant to dynamics. It was a side-comment.

Alizter

19:58

@AlexanderGruber Can you try using a different vector rather than (0,0)

Mike Miller

what? lol

Balarka Sen

confused.

Mike Miller

i quoted a section from a paper i'm reading because the "so I am told" is amusing

Balarka Sen

oh.

I thought you were being sarcastic. Nevermind.

Tien-Cheng Huang

@TedShifrin Well... Do the proofs of the Theorems in that chapter tell one how to compute?

Tobias Kildetoft

19:59

@MikeMiller Heh, nice

Transcript for

$Mathematics$ Mathematics