Mathematics - 2014-11-18 (page 2 of 4)

Integrator

16:00

@N3buchadnezzar I can't read anything except english and C++

Daniel Fischer

@N3buchadnezzar i^i is $0$ (^ = xor), try i ** i.

Wait, * has higher precedence than ^. But still, you want ** and not ^.

N3buchadnezzar

16:13

thanks

r9m

16:28

@robjohn Chris'ssis hasn't been here for more than 2 days ! :o

Hippalectryon

Still no @Chris'ssis :/

robjohn

@r9m I know. I wish she would reconsider. I think she misunderstood Pedro and so she overreacted ending up with a short suspension from chat.

Hippalectryon

@robjohn I know you're not the one who did it, but muting her on the spot wasn't a good idea either ... He could have at least warned her first ...

robjohn

All in all, I don't think it seems like a reason to give up on chat

Balarka Sen

Well, yeah, she could be annoying time to time but that's no reason to ban.

r9m

16:32

@robjohn yes I think so too .. chatting is completely harmless !! :-)

Balarka Sen

However she definitely misunderstood Pedro

r9m

@BalarkaSen I wrote the ant exam today :D

Balarka Sen

Partially because of her unfamiliarity with american expressions. "Mad" is not really meant as "mentally deranged" in it's americanized version, AFAIK. It just means angry.

@r9m Wrote?

Or sat on?

Jasper Loy

@BalarkaSen She's not annoying. She's very interesting, like me, lol.

Hippalectryon

@r9m Ant exam ??

Balarka Sen

16:34

@Jasper You're... super-non-interesting

@Hippa Analytic number theory

robjohn

@Hippalectryon well, looking at what a moderator can see about things, I think that since there were no math mods here, what was done was not completely out of order. Besides it was only a short ban from chat.

Jasper Loy

@BalarkaSen That sounds like you, lol.

Hippalectryon

@robjohn 3 hours

r9m

@BalarkaSen sat on ... :) but there was nothing much interesting/exciting though .. routine manipulations :|

Balarka Sen

Of course.

@Jasper It's a compliment for me to have someone saying that I am not interesting, so I don't mind.

:P

@r9m Any interesting question?

robjohn

16:36

@Hippalectryon I've left to work on a question for longer than that. :-) I'm not saying it isn't annoying, but I don't think it's worth leaving the site. I always thought she got something positive out of being here.

Balarka Sen

She'll come back.

Hippalectryon

@robjohn She did. even know, I don't know why she reacted so violently. I just hope she'll come back !!!

r9m

@BalarkaSen nah ,, standard textbookish stuff .. a bit lengthy calculation (perron summation shit) .. thats all ..

Balarka Sen

Oi. Perron is NOT shit.

r9m

estimating error terms suck :(

Hippalectryon

16:37

@BalarkaSen Oh as you're into NT I thought you'd be interested by this summer's IHES videos on gaps between prime numbers youtube.com/…

Balarka Sen

That's geometer-talk, @r9m.

@Hippa Been there done that

robjohn

@Hippalectryon I think that there must be some history between her and Pedro that I don't know about for the reaction that happened. At the very least, I don't think she understood what he was saying.

Hippalectryon

If she were here we could ask >.>

Balarka Sen

It's Tao's right?

@robjohn Well she misunderstood a plenty of stuff we said.

It's not her fault of course. But I guess she could do thinking that she might be misunderstanding and act a bit less violently.

Hippalectryon

@BalarkaSen What do you mean ?

r9m

16:40

@BalarkaSen I got a different interesting problem though ! estimating average order of the function $f(n) = \max\limits_{p^\alpha || n} \alpha$ :) seems nice :D

robjohn

@r9m averaging over what?

Balarka Sen

@Hippa She mostly thinks that her interpretation of our statements are right and acts violently towards them, whereas (because she is not quite adept at English, so am I, so it's not her fault) the interpretation is not quite right.

@r9m Hmm.

r9m

@robjohn estimating the summatory function of $f$ .. $\sum\limits_{n \le x} f(n)$ ..

Balarka Sen

That looks like the p-adic norm.

robjohn

@r9m better than the Euler Maclaurin Sum Formula?

Balarka Sen

16:42

@robjohn Sure.

Wait

How do you propose to use EM here?

r9m

@robjohn oh ! but how do I use EM here ? :O

Balarka Sen

JINX

Hippalectryon

@robjohn What is the dimension of $\mathcal{S}_n(\mathbb{R})$ ?

robjohn

@r9m you said you were estimating $\sum\limits_{n\le x} f(n)$

Balarka Sen

$f$ being $\max_{p^\alpha || n} \alpha$

Otherwise for general $f$ Perron is far better than EM, though far tedious.

robjohn

16:44

@BalarkaSen Oh... okay. I didn't know what $f$ was

@BalarkaSen that depends on what "general" means

Balarka Sen

well if $f(n)$ grows like $O(n)$...

and the corresponding L-function has a good complex analytic representation

robjohn

@BalarkaSen is $p$ fixed?

Balarka Sen

Yes, it's a prime.

r9m

@BalarkaSen bringing it down to perron requires atleast finding a D-series for it .. I couldn't apply the Tauberian theorems either ,, I am more or less blank :(

Balarka Sen

$f(n)$ is the largest $\alpha$ such that $p^\alpha$ divides $n$.

@r9m I hate Tauberian theorems.

Confuzzles me.

Have you inspected the arithmetic properties of $f$, @r9m?

r9m

16:48

@robjohn no no .. we are looking at the maximum of the powers in the prime factorization of $n$ .. $n = \prod\limits_{i=1}^{k} p_i^{\alpha_i}$ .. then $f(n) = \max\limits_{i=1}^{k} \alpha_i$

Balarka Sen

Oh OK

r9m

@BalarkaSen don't have much clue ..

Balarka Sen

I misunderstood it too :P

Formalize your notations dude

r9m

ah sorry .. should have mentioned we are looking at all prime factors on $n$ .. rather than a fixed prime ..

robjohn

@BalarkaSen Shouldn't that sum be somewhere close to $\frac{px}{(p-1)^2}$?

Balarka Sen

16:51

no idea

robjohn

@r9m oh, so $p$ is not fixed

Balarka Sen

@r9m $f(n^k) = kf(n)$

r9m

@robjohn yes ... sorry for the confusing notation ..

robjohn

@r9m so it's the greatest power of any prime that divides $n$

Balarka Sen

you might want to try looking at $f(nm)$ for coprime $m$ and $n$.

r9m

16:53

@robjohn yes !!

robjohn

@BalarkaSen that would just be the max of $f(m)$ and $f(n)$

r9m

@BalarkaSen thats just max{f(m),f(n)} right ?

Balarka Sen

yes, right.

hmm

$n < p^{\omega(n) \cdot f(n)}$. Does that even help?

We don't know $p$.

r9m

@BalarkaSen nah !! too large

robjohn

you'll probably need something deeper than I am familiar with to handle this.

Balarka Sen

16:57

$n < n^{1/2\omega(n) f(n)}$

that gives a dumb upper bound, right?

bah

Hippalectryon

@robjohn Have you seen my question ? :-)

robjohn

@Hippalectryon which one? your most recent one has an accepted answer

Hippalectryon

@robjohn Not that

23 mins ago, by Hippalectryon

@robjohn What is the dimension of $\mathcal{S}_n(\mathbb{R})$ ?

robjohn

@Hippalectryon what is $\mathcal{S}_n(\mathbb{R})$?

Hippalectryon

Symmetric matrices of size $n$

robjohn

17:07

@Hippalectryon over $\mathbb{R}$ it should be $\frac{n(n+1)}2$, I think

Hippalectryon

How do you show that ?

It would answer my last unanswered question on main

robjohn

@Hippalectryon write out a basis, show that for each element in the upper triangular part (including the diagonal), there is a basis element with a 1 there and a 1 in the symmetric spot

Hippalectryon

@MikeMiller ^

@robjohn I don't know any obvious basis of the symmetric matrices

@robjohn Oh wait -___-

It's quite logical

Since the lower half is determined by the upper half

robjohn

@Hippalectryon yes. Now just count the number of upper triangular positions.

Hippalectryon

Thanks ! It gives an easy answer to math.stackexchange.com/questions/1025189/…

Since nilpotent matrices can't be symmetric

robjohn

17:14

@Hippalectryon The dimension of the anti-symmetric matrices is $\frac{n^2-n}2$

@Hippalectryon not all of them are nilpotent, though

Hippalectryon

@robjohn What's the link ? (if any)

Jasper Loy

@robjohn Some are impotent.

robjohn

@Hippalectryon same argument

Hippalectryon

The dim of the sym matrices gives us an upper bound on the dim of nilpot matrices

Oh ok

@robjohn Oh wait that troubles me.

robjohn

@Hippalectryon any matrix can be written uniquely as the sum of a symmetric and anti-symmetric matrix

Hippalectryon

17:15

@robjohn In an antisym matrix, isn't the second half determined by the first one too ?

robjohn

@Hippalectryon yes, but the diagonal has to be $0$

Hippalectryon

Oh ok

Makes sense

Sawarnik

................................................................................‌.................................................................................‌..............................

Balarka Sen

Spam 2 o'clock.

Hippalectryon

@robjohn Is that right then ? math.stackexchange.com/a/1027845/150347

Swapnil Tripathi

17:19

@robjohn Is Z_p[x] an algebraic extension of Z_p?

Balarka Sen

@SwapnilTripathi No.

Swapnil Tripathi

@BalarkaSen ok! Why so?

Balarka Sen

How do you define an algebraic extension?

Recite me the definition.

robjohn

@Hippalectryon that gives you an upper bound, but not necessarily the dimension.

Hippalectryon

@robjohn Why ? I gave a space which has that dimension

robjohn

17:20

@Hippalectryon did you? I was just looking at the argument in the answer.

Hippalectryon

I used the upper triangular matrices

Strictly upper

Swapnil Tripathi

@BalarkaSen: A field E is said to be algebraic extension of F if F is a subfield of E?

Balarka Sen

Er. No, not really.

robjohn

@Hippalectryon Did you supply $\frac{n^2-n}2$ independent matrices that are nilpotent? I don't see them in either the question or the answer.

Balarka Sen

@SwapnilTripathi The correct definition is that E over F is an algebraic field extension if F is a subfield of E and every element in E is algebraic over F, i.e., is a root of a polynomial over F.

Hippalectryon

17:23

@robjohn Isn't the dimension of $T_n^{++}$ $\dfrac{n(n-1)}2$ ?

robjohn

@Hippalectryon what is $T_n^{++}$?

Hippalectryon

Strictly upper triangular matrices

Balarka Sen

@SwapnilTripathi Can you give me an example of an algebraic extension?

Swapnil Tripathi

Q(\sqrt{2}):Q

Ok! Man, I was on the wrong track all this time!!! So Q(\pi):Q is not algebraic extension?

robjohn

@Hippalectryon Okay, then I guess so.

Hippalectryon

17:24

@robjohn Just by taking the $n(n-1)/2$ elementary matrices $E_i,j$ with $i<j$

@robjohn ok thanks

Balarka Sen

@SwapnilTripathi OK. Why is Q(\sqrt{2})/Q algebraic?

And why is Q(\pi)/Q not?

robjohn

@Hippalectryon I just wanted to make sure that everything was covered in the answer or the question.

Swapnil Tripathi

I understand what you mean. :) f(x)=x^2-2. \pi is transcedental, so no such polynomial.

Hippalectryon

@robjohn Well, R. Israel gives some details in the comments

Swapnil Tripathi

@BalarkaSen ^^

Balarka Sen

17:26

@SwapnilTripathi I am asking you to show a proof of the (correct) fact that Q(\sqrt{2}):Q is algebraic and Q(\pi):Q.

f(x) = x^2 - 2 is a polynomial, not a proof ;)

Swapnil Tripathi

That will lead to the proof, right? @BalarkaSen

Hippalectryon

@BalarkaSen LaTeX plz :c

Balarka Sen

Sure, @SwapnilTripathi. But complete the proof.

robjohn

@Hippalectryon no, he just mentions that the upper triangular matrices are nilpotent. He doesn't go much further.

Hippalectryon

@robjohn I mean, he gives their dimension

Balarka Sen

17:30

I'm lazy @Hippa

Hippalectryon

@BalarkaSen I know

@BalarkaSen Go eat some infants

Balarka Sen

and their dessins

robjohn

@Hippalectryon well, yes, that is true, and they do form a subspace, but he doesn't show that that is all nilpotent matrices

Hippalectryon

@robjohn mhm true

@BalarkaSen ????

Rafflesia arnoldii

Is it just me, or is the letter spacing in this post horribly wrong? Reminds me of xkcd1015.

Balarka Sen

17:31

@Hippa You're supposed to be a french. Dessin means drawing.

Hippalectryon

@BalarkaSen infants don't draw

@Rafflesiaarnoldii It's weird indeed

Balarka Sen

@Hippa I was making a pun at the mathematical theory of dessin de'nfants found by Grothendieck.

Swapnil Tripathi

Q(\sqrt{2}):Q={1,\sqrt{2}). All elements are of the form a+b\sqrt{2}. We need to show this is algebraic over Q. The (x-a)^2/b^2-2 gives a polynomial over Q with root a+b\sqrt{2} @BalarkaSen

Hippalectryon

@BalarkaSen oh

Balarka Sen

That works, @SwapnilTripathi. And the logic that \pi is transcendental works for showing that Q(\pi):Q is not algebraic.

dorothy

17:34

hi @robjohn

Balarka Sen

@SwapnilTripathi So now do you see why Z_p(x) is not algebraic over Z_p?

Swapnil Tripathi

Yes, I do. :) @BalarkaSen. I was thinking about whether Z_p[x] was a perfect field and ventured too deep and too wrong! :D Any help on that?

Hippalectryon

@robjohn Adding two nilpot matrices doesn't make a nilpot matrix ? Let $A,B$ two nilpot matrices with $A^a=B^b=0$, and $m_n=\max\{\binom{n}{k}\mid1\le k\le n\}$, then $(A+B)^p=\sum\binom{n}{k}A^kB^{n-k}$ hence if $m_p\ge\max(a,b)$ then or $A^k=0$ or $B^{n-k}=0$. Am I wrong ?

robjohn

@Hippalectryon see my second comment

Balarka Sen

@SwapnilTripathi Z_p[x] is not even a field.

dorothy

17:37

@robjohn I think you need n = i when for n = 6, 10, 14 and n = i-1 for n = 4, 8, 12

Swapnil Tripathi

How can we show that? @BalarkaSen If it has a simple proof?

Hippalectryon

@robjohn Why doesn't the space of symmetric matrices form a subspace of dimension (n^2+n)/2 ?

robjohn

@Hippalectryon consider $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\1&0\end{bmatrix}$. Each is nilpotent, but their sum is not

Swapnil Tripathi

I read it somewhere that infinite fields are perfect. I think they should have meant fields of char 0 @BalarkaSen

robjohn

@Hippalectryon I said it does form a subspace

Swapnil Tripathi

17:38

That got me confused

Hippalectryon

@robjohn I can't read sorry -_____-

damn me

Balarka Sen

@SwapnilTripathi car 0 fields are perfect yes.

trivially so, in fact

Swapnil Tripathi

@BalarkaSen I know that. But they shouldn't loosely say that infinite field are perfect. :/

Considering the fact Z_p[x] is not

Balarka Sen

yes, right, that is not true. not all infinite fields are perfect.

@SwapnilTripathi Z_p[x] is not a ~~good~~ counterexample.

it's not a field. just a commutative ring.

mathh

Hey everyone. I think this wikipedia article has a little error. It should be "If k >= 2, then a is called a multiple root.", not "If k > 2, then a is called a multiple root."

Hippalectryon

17:42

@robjohn I understand, but I don't see where my proof is wrong. Can you point out the error ?

Swapnil Tripathi

ok! Z_p(x)? @BalarkaSen

Balarka Sen

mmhmm

Swapnil Tripathi

@mathh Yes k>=2?

@BalarkaSen: That's a yes?

Balarka Sen

yes.

mathh

@SwapnilTripathi I said it should be "k>=2". What's your question?

Swapnil Tripathi

17:44

@mathh: I'm telling you that youre correct in saying that

Balarka Sen

wiki is full of typoes

mathh

@SwapnilTripathi Well there's a question mark in the end.

Swapnil Tripathi

@BalarkaSen: Can you share a link to the proof? If it is not very tough?

Balarka Sen

@SwapnilTripathi it's not tough. how do you define perfectness?

Swapnil Tripathi

@mathh Oh sorry. I was too consumed in my doubts. @BalarkaSen will understand. :D

There are some equivalent conditions. I remember only a few. First is frobenius map is bijective?

Balarka Sen

17:47

right.

Swapnil Tripathi

Second is, perfect iff every algebraic extension is separable.

Balarka Sen

can you use the first one?

Swapnil Tripathi

@BalarkaSen First will do?

Balarka Sen

mmhmm

Swapnil Tripathi

Coefficients can be written as p^th powers. Don't know about the x's

When we define Z_p(x), does x have a domain? @BalarkaSen

Balarka Sen

17:48

well obviously you need to look at the K-transcendental x.

@SwapnilTripathi domain?

r9m

I need some sleep ... BBL :-)

Swapnil Tripathi

x is a variable, isn't it? :-O @BalarkaSen

Balarka Sen

don't let the L-functions bite @r9m

r9m

@BalarkaSen ;) okay ... gn

Balarka Sen

@SwapnilTripathi not sure what you mean by variable. it's just a K-transcendental.

Swapnil Tripathi

17:51

@BalarkaSen I used to think these are functions like those in real analysis! :D

Mike Miller

@Rafflesiaarnoldii Poor Martin-Blas is taking the bait. Now he's going to be harassed by Thierno until the original post is deleted.

Swapnil Tripathi

I sure suck!

Balarka Sen

@SwapnilTripathi heh?

you realize you're speaking nonsense, no?

Swapnil Tripathi

Yes, I do. Long day.

So going back to x!!! What do I need to do? @BalarkaSen

Ted Shifrin

hi @Balarka @Mike @r9m @Swapnil

Swapnil Tripathi

17:54

Hi @TedShifrin :D

Balarka Sen

you need to show that there is an element a in Z_p(x) such that t^k = a has no solution in Z_p

@TedShifrin!

r9m

@TedShifrin hello professor !! :) .. I was about to go to sleep .. tc gn :-)

Ted Shifrin

g'night, @r9m

robjohn

@r9m Oh no... good night

Ted Shifrin

hi @robjohn

Hippalectryon

17:57

@robjohn Nvm my last message

Ted Shifrin

oh no, @Hippa is here

Hippalectryon

@TedShifrin (҂ಥ◇ಥ҂)

robjohn

@Hippalectryon yes, the binomial theorem assumes you are in a commutative ring :-)

r9m

@robjohn ? :-)y u say oh no ?

robjohn

@TedShifrin good morning

Swapnil Tripathi

17:59

x has no solution of such form because if it had a solution then t^p=x and x will not transcedental as it would become a root of a-t^p=0? I don't know. :| @BalarkaSen

Hippalectryon

@TedShifrin Be happy, I'm off for dinner

robjohn

@TedShifrin or is it afternoon there now?

@r9m you are leaving... I was away on mod duties and missed out on your presence :-)

Swapnil Tripathi

Hey @hippa :D Forgot to greet you.

Balarka Sen

@SwapnilTripathi you're more-or-less right.

Ted Shifrin

à bientôt, @Hippa

it's 1 PM, @robjohn

Balarka Sen

18:02

it's barely afternoon in this part of the world.

w.r.t me, of course :P

Ted Shifrin

@Balarka: You seem to live in time reversal.

Swapnil Tripathi

@BalarkaSen: But you're in W.B., right? :D

Balarka Sen

sure.

evinda

Hello @BalarkaSen!!! :-) Long time, no see..

Balarka Sen

@TedShifrin well, obviously. homotopy equivalence is symmetric.

(pun on "time reversal")

Swapnil Tripathi

18:04

@BalarkaSen: More or less correct. How do I push it to being absolutely correct?

Ted Shifrin

stop flaunting your meager topology knowledge, @Balarka :D

robjohn

@TedShifrin Ah, time for some tea here... I just need to select a variety :-)

r9m

@robjohn :D I am only taking a nap for a couple of hours ;) ... gn thank you :)

Balarka Sen

@TedShifrin I have just began studying algebraic topology.

Ted Shifrin

make it algebraic, @robjohn

robjohn

18:04

@r9m sleep well (but not in a well)

Balarka Sen

Barely know what a fundamental group is.

Ted Shifrin

Mine too was a pun, @Balarka

S.C.

hello

Balarka Sen

@TedShifrin oh?

Ted Shifrin

meager sets, @Balarka

hi @S.C.

Mike Miller

18:05

You should be ashamed for that, @Ted

Balarka Sen

Oh LOL

r9m

@robjohn haha lol !! I'll do well by keeping away from a well then, lest I slip into a well and finally sleep well .. :P

Ted Shifrin

no, @Mike, but you would be.

Balarka Sen

why is it that most users in MO write shitty LaTeX?

S.C.

Emm. I have a problem in hand which i am struggling with. Let H be a hilbert space and let ${u_{n}}$ be an orthonormal basis. Let {v_n} be a orthonormal subset of H such that $\sum_{n \geq 1} ||u_{n}-v_{n}||^{2} < 1$. Then i have to show {v_{n}} are also an orthonormal basis. Any ideas>>

Balarka Sen

18:11

@Mike I want to prove that the two maps $S^1 \to S^1$ defined by sending $x$ to $x$ and sending $x$ to $-x$ resp are homotopic. Any hint?

Mike Miller

No.

Balarka Sen

I have proved that if $f : S^1 \to S^1$ is not homotopic to the identity map, then $f(x) = -x$ for some $x$ in $S^1$.

@MikeMiller Let the old ~~grudges~~ gauges be forgotten.

rehband

@DanielFischer Hey Daniel! You were saying $S^2/\sim \cong D^2$, right? (where $x \sim -x$ )

Mike Miller

@rehband No, not at all. $S^2/\sim \cong \Bbb{RP}^2$.

Hippalectryon

@TedShifrin Sorry to disappoint you, but I'm back ＼(⊙㉨⊙ ＼ )

Ted Shifrin

18:16

Merde, alors.

Balarka Sen

What the hell is $\sim \cong$?

Hippalectryon

@MikeMiller Have you seen the proof ?

Mike Miller

What you're probably thinking of is what you get by reflecting along a plane.

rehband

@MikeMiller Right

Hippalectryon

@:O THAT'S BAD WORDS >:c

Even I don't use that in France

Ted Shifrin

18:17

Well, plenty of people do, @Hippa.

Balarka Sen

what does that even mean?

Hippalectryon

@BalarkaSen S**t

Balarka Sen

clearly that's not bad

Hippalectryon

@TedShifrin Still it's bad :/

@BalarkaSen -_____-

Ted Shifrin

Alors, tu ne me parleras plus, @Hippa?

Balarka Sen

18:18

@Hippa You mean Suit, don't you?

How is that supposed to be bad?

Mike Miller

@Rehband If you let $(x_1,x_2,x_3)\sim (x_1,x_2,-x_3)$, then $S^2/\sim \cong D^2.$

Hippalectryon

@BalarkaSen Of course not. I mean sect :c

:DDDD

@TedShifrin Oh si, je vais encore avoir besoin d'aide ;-)

rehband

@MikeMiller Ok yes, I see

Balarka Sen

@Ted is trying to express that his suit has become too tight for him.

He needs to lose some weight.

Hippalectryon

@BalarkaSen That's rude too :c

Ted Shifrin

18:20

@Balarka: Ted has never owned a suit ... but, yes, Ted should lose weight.

Hippalectryon

@BalarkaSen that was rude for the suit :D

Balarka Sen

LOL @Ted

Hippalectryon

Mike Miller

@Ted I just got my results back; I passed my French test. I may be done with tests for the rest of my life. :P

Ted Shifrin

Maybe not @Mike.

Balarka Sen

18:23

French ~~test~~ toast?

Mike Miller

Don't say that, @Ted.

Balarka Sen

@Mike C'mon. Give me a hint on that one.

robjohn

@S.C. about the only thing left to show is that the orthonormal set spans the space

Mike Miller

It is seriously easy. I wouldn't give someone a hint on showing that $\Bbb Z_2 \oplus \Bbb Z_2$ is abelian. Use your brain and you'll solve it quickly.

Hippalectryon

@TedShifrin I was wondering why do you record your classes ?

Balarka Sen

18:27

@Mike OK.

Hippalectryon

Why is Right Hand rule nonsense ? :c

Balarka Sen

I gotta sleep. Byes.

skullpatrol

later pal

Hippalectryon

18:42

@r9m You don't have any link (email etc) to @Chris'ssis either, do you ?

@robjohn ?

robjohn

@Hippalectryon I only have what I know from being a mod, and cannot give that information out.

Hippalectryon

@robjohn Oh i see :/

@robjohn ~~Do you accept bank notes ?~~

Math corruption

Daniel Fischer

@rehband No, I'm saying that the (closed, of course) disk is homeomorphic to a closed hemisphere. And you get the projective plane by identifying antipodal points on the equator of the hemisphere. Or, equivalently, identifying antipodal points on the boundary of the disk.

rehband

@DanielFischer Yep, I got it now. Thanks

skullpatrol

@Hippalectryon she'll come back, eventually, as Daniel said...

Hippalectryon

18:48

@skullpatrol I hope so :/

@skullpatrol Usually she came back faster

skullpatrol

@Hippalectryon me too :\

@Hippalectryon she's probably very busy

+ christmas is around the corner ...

Hippalectryon

@robjohn Since you're here, it'll save a question on meta : since we have quite a lot of questions / day, woudn't it be doable to have customized newsletters ? The current one is the same for everyone, I'm pretty sure that being able to receive the most popular Q/A of the week in one's fav categories would be great

@skullpatrol Lol

No Christmas for maths :)

With maths it's Christmas everyday !

skullpatrol

:D

Everyday is a gift, my friend.

Jasper Loy

Sometimes, it is a curse.

But what is good and what is bad, that is a very deep question.

The good can become bad, and the bad can become good.

skullpatrol

its all about perspective

Jasper Loy

18:59

I just read the story of Alan Turing. Seems the British laws were sick back then.

It is interesting to see how the world evolves. Some first, others later. Some quickly, others slowly.

skullpatrol

yes, the "war act" was in full effect back then

Jasper Loy

I am talking about the homosexual act, lol.

skullpatrol

so am i pal

Jasper Loy

OK, lol.

skullpatrol

:-)

robjohn

19:03

@Hippalectryon You'd have to suggest that on meta as a feature-request. I can't do anything about it beyond what you could with a meta post

skullpatrol

@Hippalectryon sounds like a great idea

+1 from me

Daniel Fischer

More Thierno crankdom. If anybody feels like making it insta-deletable, don't hesitate.

3

Hippalectryon

0

Q: Customized newsletter

Being a mathematics website, the questions asked here are on pretty diverse areas. However, the weekly newsletter is the same for everyone. I believe that being able to receive customized newsletters, i.e., receiving the most popular Q/A of the web in one's favorite categories/tags would be fair...

feature-request

UserX

@Hippalectryon I don't think that's a good idea...

evinda

19:18

@DanielFischer Hi!!! I want to show that $I(V(y-x^2,z-x^3)) \subseteq <y-x^2,z-x^3>$.

Let $f(x,y,z) \in I(V(y-x^2,z-x^3))$. Then, $f(x,y,z) \in \mathbb{C}[x,y,z]$ such that $f(t,t^2,t^3)=0$.

Then, we apply the euclidean division of f with y-x^2.

$$f(x,y,z)=f_1(x,y,z)(y-x^2)+f_2(x,y,z)$$

If we have a(x)=b(x)q(x)+r(x), then it is: $deg(r(x))< deg(b(x))$

How is it now where we have a multivariable polynomial? How can we find $deg(f_2(x,y,z))$?

UserX

The site is strictly to give answers to questions about mathematics.

Hippalectryon

@UserX What about it ?

dorothy

anyone got any ideas about my question math.stackexchange.com/questions/1023351/… ? Sorry to spam but it is feeling lonely

2

UserX

Let $\Bbb R^1$ be our metric space. How do we prove that every $x\in\Bbb R$ is a limit point of $\Bbb Q$

Jasper Loy

19:33

@dorothy It can't be more lonely than I am.

UserX

I lost my consecutive streak :(

And I logged in for a lot of time each day, I just wasn't active...

Hippalectryon

@JasperLoy You're not lonely, we're here !

Jasper Loy

@Hippalectryon Thank you.

Daniel Fischer

@evinda In that situation, view $f$ as an element of $(\mathbb{C}[x,z])[y]$. Then since $\deg_y (y-x^2) = 1$, you have $\deg_y f_2 = 0$, that is, $f_2 \in \mathbb{C}[x,z]$, if you view that as a subring of $\mathbb{C}[x,y,z]$. In the next step, use $\deg_z (z-x^3) = 1$.

UserX

skullpatrol

19:47

lol^

UserX

Jasper Loy

@UserX I have seen this 9000 times.

UserX

9001 it is then :D

dorothy

I am not sure it is suitable for mathoverflow

Jasper Loy

@dorothy That site is already overflowing.

dorothy

19:56

@JasperLoy sorry to hear it..

@JasperLoy also I feel pretty sure that someone competent could write down a formula for it

@JasperLoy maybe not a pretty one, but at least a formula

which I think means it's not relevant to mathoverflow

I am not sure why it hasn't attracted the interest for smart math.SE people

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$Mathematics$ Mathematics