Mathematics - 2014-12-15 (page 3 of 10)

That's pretty cool actually

@KajHansen Pedro discouraged us :P

What did he say?

10:10 AM

"That's not gonna work" or something like that :P

LOL. There are tons of things that shouldn't work but do.

I give it to you to think about @Kaj. You are more capable of coming up with something useful than me, seeing that you are much more natural with group theory than me.

Sure. I have lots of time over this break. It's awesome.

Haha, cool. Do let me know if you come up with something.

More or less the objective is this : $G$ and $G'$ be groups with given isomorphic galois groups $Gal(F/\Bbb Q)$ and $Gal(F'/\Bbb Q)$. Then construct a field extension $E$ of $\Bbb Q$ given a group extension $K$ of $G$ and $G'$ such that $K \cong Gal(E/\Bbb Q)$

^ @Kaj

By $K$ is a group extension of $G$ and $G'$, I mean there is a short exact sequence $1 \to G \to K \to G' \to 1$, or in other words, $K/G \cong G'$

Lembik

10:29 AM

hi.. is this OT for math.se ? stackoverflow.com/questions/27474181/…

Getting lazy with the LaTeX again I see :)

There. :P

@KajHansen Can we do this for $1 \to A_4 \to S_4 \to \Bbb Z_2 \to 1$?

Quickly, give me an extension corresponding to A_4.

I'm not read up on short-exact sequences @BalarkaSen. Another thing I'll need to study this break.

@BalarkaSen $$\space \int_0^{\infty} \left(\frac{1}{\sqrt[3]{3x+1}}-\cos(x^2)\right)\frac{1}{x} \ dx$$

@KajHansen I mean that I want to do it for group A_4 and Z_2 and construct S_4 from these two.

10:33 AM

Ups ... I forgot ...

S_4 is the semidirect product of A_4 and Z_2, not?

Sure, I'm with you

So give me a number field corresponding to A_4.

So by an extension, you are equivalently asking for a group $K$ such that $K/A_4 \cong ?$

no, i am asking for a gal ext K of Q such that Gal(K/Q) \cong A_4

10:39 AM

Hmm, let me think for a sec.

just to clarify, i don't know one off the top of my head either :P

Why is it simple to think of such a group?

Such an extension I mean to say.

@KajHansen hm?

@kaj Why aren't you sleeping?

I figure A_4 is generated by a $3$-cycle together with a composition of $2$-cycles.

10:41 AM

ok, yeah

Why'd you pick $A_4$ out of all the possible groups @BalarkaSen ?

@JasperLoy, I will be soon.

well i wanted to try the galois theoretic analysis for semidirect product of nontrivial groups

and S_4 = A_4 \ltimes Z_2 is the one that springs to mind

Oh, I see why

I'm trying to construct this extension by first finding a polynomial with such a Galois group.

Any polynomial with $4$ complex roots will have the composition of $2$-cycles. I'm trying to also construct one that includes the $3$-cycle.

ok, idea : galois group of Q(2^{1/4}, i) over Q is D_4. so you need a degree 3 extension of Q(2^{1/4}, i).

hrm

That can't be too bad. Does it have to be Galois over Q?

Oh I guess it does.

10:49 AM

yes

i googled, but ended up with weird extensions @KajHansen :(

:/

OK, we are just goofing off

Let Gal(K/Q) \cong A_4, Gal(K'/Q) \cong Z_2

How are $K$ and $K'$ related?

We want to find an E such that Gal(E/Q) \cong S_4

@KajHansen they're not.

k

10:58 AM

OK, so if we have such an E, then Gal(E/Q)/Gal(K/Q) \cong Z_2

But fundamental theorem says Gal(E/Q)/Gal(K/Q) \cong Gal(E/K)

So E is quadratic over K.

I'm not sure it'll be any easier to find such an E.

You mean K

E is not very hard to find. I think there is a theorem that says x^n - x - 1 has galois group S_n for all n

Oh that's pretty cool. I'd like to read that proof later.

The inverse Galois problem fascinates me.

indeed, but IGT for S_n is way way easier than determining an explicit polynomial over Q that has galois group S_n

it's called hilbert's irreducibility theorem, what you want

says that if you have a group which is realizable over Q(z), then it's also realizable over Q

It's like 6 AM here, so I really need to go to bed soon. I'll be back tomorrow ofc

skullpatrol

11:07 AM

Later pal

ok, bye @Kaj

I just put on a hat.

Khallil Benyattou

11:24 AM

All you gotta do is click "Show Controls" in order to change the size of and rotate the hat, @Balarka!

Then you're all set to be, in the immortal words of @Kaj, "gangsta". ^_^

Done, but not exactly "gangsta"

It rather looks like someone threw a birthday cake on my face.

skullpatrol

LEL

11:45 AM

@Chris'ssis Here is what I did:

Note that
$$
\begin{align}
\int_0^\infty\frac{(1+ax)^{-p}}{x^t}\mathrm{d}x
&=a^{t-1}\int_0^\infty\frac{x^{-t}}{(1+x)^p}\mathrm{d}x\\
&=a^{t-1}\frac{\Gamma(1-t)\Gamma(p+t-1)}{\Gamma(p)}\\
&=\frac{a^{t-1}}{1-t}\frac{\Gamma(2-t)\Gamma(p+t-1)}{\Gamma(p)}\tag{1}
\end{align}
$$
Furthermore,
$$
\begin{align}
&\lim_{t\to1^-}\left[\frac{a^{t-1}}{1-t}\frac{\Gamma(p+t-1)}{\Gamma(p)}-\frac1{1-t}\right]\\
&=\lim_{t\to1^-}\left[\frac{a^{t-1}-1}{1-t}\frac{\Gamma(p+t-1)}{\Gamma(p)}\right]+\lim_{t\to1^-}\left[\frac{\Gamma(p+t-1)-\Gamma(p)}{(1-t)\Gamma(p)}\right]\\[9pt]

Anonymous

@JasperLoy I think I am going to change my branch from Engineering to Mathematics just as you wanted.If I was rude to you before,it's because my English sucks!

@robjohn Nice :-) It's good to know more approaching ways!

@robjohn did you see my way?

@Chris'ssis yes, but I haven't gone through it all yet.

@robjohn OK

@Chris'ssis Ah, yes, I see you used the integral for $\gamma+\psi(p)$.

11:55 AM

@robjohn Yeap.

@Chris'ssis I would need to verify that, since it is not one that I have seen before.

@robjohn Which one to verify?

Digamma function

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...

@robjohn It's straightforward by geometric series.

@Chris'ssis If you link to the specific section‌, it makes things easier.

@robjohn By the way, how can I do that? Sometimes it doesn't work. I meant to do that.

@Chris'ssis Each section on Wikipedia has a separate link. They are given in the Contents box. Copy one of those links.

12:02 PM

@robjohn Right.

@Chris'ssis In any case, I am not denying it is true, I would just need to verify a formula I had not seen before. That's all.

@robjohn OK. If you remember I proposed in the past a question that used that representation.

I think it was about this one $$\int_0^1 \left(\frac{1}{1-x}+\frac{1}{\log(x)}\right)^2 \ dx$$

ADG

12:29 PM

Some resource for non elemetary functions?

@Ashwin Just think carefully first, and then act according to your plan. Good luck!

@Ashwin No need to apologise, you were not rude and I am only a banana.

Everyone uses hat. Except @robjohn. Come on, we should celebrate Winter Bash. Feel the joy!

evinda

Hello @DanielFischer!!!

@evinda May I ask why you only ask Daniel for help and not the others?

Sheesh. I got a secret hat.

evinda

12:41 PM

@JasperLoy We began talking about an algorithm and I wanted to ask him if he has time so that we continue with the discussion... :)

@evinda OK. Anyway, I think most of the time it is better to post your question on the main site.

4

I don't post my questions on the main @Jasper

I think chat is more for questions that don't fit well on the main site, like those that involve discussion and not definite answers.

Oh, sure.

I agree.

Also, chat is for losers like me to talk about my life problems.

12:44 PM

The questions I pose are either very brilliant or very trivial.

@BalarkaSen You seem to be studying algebraic instead of general topology now.

Someone is starring just to get a hat, it seems.

@JasperLoy Yes.

I am.

I will be studying calculus in 17 days.

@Venus @robjohn does not have any hat!

@BalarkaSen Everyone is starring things in the Eng room as well.

12:47 PM

@Venus Seems like he chose I hate hats

Star the already starred posts instead of nonsense.

11 hours ago, by Alexander Gruber

(WB hat list) If you just want Sumo Judge, please add stars to these, not random posts.

@JasperLoy and everyone is down-voting posts!

@Integrator I resolved not to downvote anything ever on my current accounts.

You have more than one, @Jasper?

@BalarkaSen Well, I mean on math and Eng main and meta, that is 4.

12:49 PM

ah

@Integrator I like this funny hat. I plan to use my real photo as avatar ^^

@JasperLoy You're a clean man!

@Integrator I am not very clean, as my mother would tell you. Sometimes, I brush my teeth only once a week.

I never brush.

2

@Venus Go on! I want to see a fresh woman ;)

12:51 PM

I've done 3 downvotes today, why does my rep not decrease? A bug?

@BalarkaSen What do you mean?

That I never brush my teeth ;)

@Venus You only lose a point when you downvote an answer, not a question.

It's true.

@Venus On answers? Check if those answers are deleted!

12:51 PM

@BalarkaSen Really?

Yep.

I stink.

@BalarkaSen In your whole life?

@JasperLoy Are you serious? Why I didn't know that?

@Venus Yes. You have not been here for too long.

@JasperLoy Well the subset of days I brushed in my life is a sparse subset of my life.

12:53 PM

@BalarkaSen OK. I can tell you that my teeth are in good condition though they are stained.

@JasperLoy What happen to the user that the question I downvote? Does his/her rep decrease?

@Venus Anyone who gets downvoted loses 2 points.

That's fun! I'll start downvote everyone's questions then, hahaha

@venus Quick put up your picture!

@Venus Don't forget to ping me when you're done.

12:57 PM

aw hell. everyone is starring everything.

It is well known that I never star anything in chat.

stupid hat.

Khallil Benyattou

Why is everybody starring everything?

Stars are everywhere ^^

@Venus Are you a engineering student?

1:12 PM

@Integrator Not yet, almost

@Venus almost?

@Integrator Ask @Huy ^^

@Venus :0 Huy?

@Venus Fine!

@Chris'ssis The integral should be the negative of what you have in the image:

$$
\begin{align}
\int_0^1\frac{1-x^s}{1-x}\mathrm{d}x
&=\lim_{t\to1^-}\left[\int_0^1\frac1{(1-x)^t}\mathrm{d}x-\int_0^t\frac{x^{s}}{(1-x)^t}\mathrm{d}x\right]\\
&=\lim_{t\to1^-}\left[\frac{\Gamma(1)\Gamma(1-t)}{\Gamma(2-t)}-\frac{\Gamma(s+1)\Gamma(1-t)}{\Gamma(s+2-t)}\right]\\
&=\lim_{t\to1^-}\frac{\Gamma(2-t)}{1-t}\left[\frac{\Gamma(1)}{\Gamma(2-t)}-\frac{\Gamma(s+1)}{\Gamma(s+2-t)}\right]\\
&=\lim_{t\to1^-}\frac{\Gamma(2-t)}{1-t}\left[\frac{\Gamma(1)-\Gamma(2-t)}{\Gamma(2-t)}-\frac{\Gamma(s+1)-\Gamma(s+2-t)}{\Gamma(s+2-t)}\right]\\

3

@robjohn Yeah. Isn't this way a bit too difficult? I mean using geometric series is a fast, easy way.

1:24 PM

@Chris'ssis Does geometric series work for all values of $s$?

@robjohn I don't know what you think of. That way works.

@Chris'ssis Perhaps I am not thinking of the way you are. The way I was thinking of using geometric series only works for $s\in\mathbb{Z}$.

@robjohn Not really.

@Chris'ssis Ah, you are expanding $\frac1{1-t}$

@robjohn Right.

1:34 PM

I am a mean girl, haha

@Studentmath!

@Balarka!

What's cookin'?

Topology, finally moved to topological spaces (rather than metric ones). How about you?

LEL your hat @Student

1:36 PM

It's a bit oversized, but I like it that way

@Studentmath I am thinking about a Galois theory problem in an algebro-topological way.

Thomas

@robjohn: Hey. Could you take a look at the flag I just raised on: math.stackexchange.com/questions/1067909/…

What problem? (Probably won't understand, but worth the shot)

Do you know galois theory?

Just a bit, from coding theory mostly actually

1:39 PM

OK, then I guess you know what Galois groups are?

Yes

OK, then you can definitely understand the problem : Can we have a geometric representation for $Gal(\overline{\Bbb Q}/\Bbb Q)$?

What does overbar-Q stands for?

Algebraic numbers.

Q adjoined with all the algebraics over Q.

It's sometimes written as $Gal(Q)$, right?

1:45 PM

Yeah probably.

I guess I have seen it written like that.

Interesting, does it have one? I would guess yes, but no idea.

@Studentmath Yes, there's something done by Grothendieck out there. In fact this question was posed in Grothendieck's Equisse d'un Programme.

I have an approach, but it is still just an approach :P

2:02 PM

@Chris'ssis Could you downvote this question of mine? Thanks. ^^

Where is your hat, @N3buchadnezzar?

N3buchadnezzar

@BalarkaSen W(hat)?

@N3buchadnezzar, the king of puns!

Lambert W-function evaluated at a hat blows up though @N3buchadnezzar

N3buchadnezzar

@BalarkaSen I have a cat instead of a hat, it did belong to postman pat. However when he sat, I hit him with a bat. He said what and hit the mat, and now postman pat has no cat.

How does one aquire such hats?

@N3buchadnezzar Look at the Winter Bash hat list

N3buchadnezzar

2:13 PM

Now i understand the star obsession in chat

c(hat)

user21820

hello! sorry to interrupt.. i'm having problem with chat.stackexchange when my Ad-Block addon is on.. (I had to switch it off to get in here.) Does anyone know what I've to white-list?

Alexander Gruber

Dang it pay attention to the sidebar

@AlexanderGruber!

Daniel Fischer

@AlexanderGruber I'm tempted to star that.

2:16 PM

LEL

Daniel Fischer

even more tempted

@DanielFischer I'm tempted to star that

@AlexanderGruber to the rescue.

@Integrator Don't be too surprised if math110 accepts my answer, see this for the reason. ^^

I want a hat like this

@BalarkaSen That's not a hat. It's just a shirt :D

2:20 PM

Yes, @Venus, his is wearing his nighsuit as a hat.

Hmph. sleep has dropped me to third.

@MikeMiller If that short exact sequence is holds, then $1 \to \check{\pi_1}(X) \to \check{\pi_1}(X/\mathbf Z_p) \to \mathbf Z_p \to 1$ where $X$ is the $p$-adic solenoid.

Alexander Gruber

There had better be a secret hat for canceling a crap ton of stars grumble

@MikeMiller Plus, I think $\check{\pi_1}(X) = 0$.

@Venus Why do you want me to downvote your question?

2:25 PM

So I guess $\check{\pi_1}(X/\mathbf Z_p) \cong \mathbf Z_p$ holds.

@DanielF Do you think I misinterpreted the OP of this question?

@Chris'ssis Because I want a hat :D

@Venus OK. Done. :-)

@Chris'ssis Thank you ^^

@Venus Welcome :D (thanks for being downvoted)

Daniel Fischer

2:28 PM

@MikeMiller No, that's the way I interpret the assignment too. I have however the impression that the OP (and Peter F) have misunderstood it.

@Chris'ssis I'm pleased to be downvoted because I wanna collect hats. More hats :D

OK :-)

Okay, just checking.

@Chris'ssis Wait!? Why have I not received the "Business in the front, Party in the back" hat? It seems I misinterpret the detail of this hat.

@MikeMiller Are there any fundamental group that can handle algebraic curves over projective planes? In particular, is there a fundamental group that can handle spaces with Zariski topology?

Cech looks too weak for it.

2:33 PM

Finally I get it :D

@BalarkaSen What do you want it for?

Eh, I'll spill the beans @Mike. I want to realize Gal(\bar C(z)/C(z)) geometrically. I believe I can do that in a similar way as Z_p using solenoid Riemann surface instead of just usual solenoids. I need to compute fundamental group of algebraic curves to do that.

I'm ignoring that. Why do you need to compute the "fundamental group of algebraic curves" if you don't know what it should mean? More importantly, you didn't answer the question - what should be the point of the fundamental group in this situation?

We actually not Gal(\bar C(z)/C(z)). You can still compute fumdamental group of solenoid Riemann surface obtained by taking inverse limit of all Riemann surfaces over P^1 using Cech, I think

I think I'd need it in Gal(\bar Q/Q)

You didn't answer either of my questions.

2:40 PM

@MikeMiller "Why do you need to compute the "fundamental group of algebraic curves" if you don't know what it should mean?" I believe Gal(\bar Q/Q) acts nicely on an object which is inverse limit of an inverse system of algebraic curves over \bar Q

That is why

@Venus I don't understand that!

Okay, I'm not convinced.

@Venus is math110 and chinamath same person?

"what should be the point of the fundamental group in this situation?" If G acts nicely on an object X, we should hope for a SES 1 --> pi_1(X) --> pi_1(X/G) --> G --> 1 and thus realize G geometrically.

@Integrator Did thinks so, me too.

2:42 PM

As I want Gal(\bar Q/Q) realized geometrically, I'd naturally want fundamental groups to be defined for my objects on which G act nicely.

@Venus I don't have any problem that he accepted your answer.

@Integrator Can you help me for this one.

@Integrator I know, just saying

@MikeMiller Well, you didn't really listen to all the stuff I have said about realizing Z_p geometrically using solenoids. I believe similar should work for Gal(\bar Q/Q)

I was really using Z_p version as a prototype.

It's the simplest "interesting" profinite group you can think of.

That's not what the point of a fundamental group is, that's what you want to use it for. A fundamental group is some sort of invariant for the objects in your situation; usually homotopy invariant. (I don't know how to talk about homotopy invariance for varieties - maybe birational equivalence?) The only thing I know of is the etale fundamental group.

W. A. T.

2:45 PM

last I heard you hadn't done anything successfully there...

Etale stuff is about that?

@MikeMiller I'm close.

I have to prove that the Cech fundamental group of the solenoid is trivial

Then the rest would follow.

k

@Venus I'm almost done with that just putting all pieces together $$\frac{-2 \sqrt{-x^4+x^2+2}+\left(3-6 x^2\right) \log \left(2 \sqrt{-x^4+x^2+2}+3\right)+3}{24 x^2-12}$$

@Venus Don't edit your question, before I post my answer.

Can't wait. If I may ask, please provide the details answer. ^^

I won't :-)

@Venus That was indefinite integral, and the result is $$\frac{1}{6} \left(3-2 \sqrt{2}\right)$$

Anonymous

2:48 PM

@JasperLoy Do you really think what I am doing is right?
Because I got huge opposition from friends and family.

@Venus I said I'm almost done! Will post my solution soon.

@Integrator I have put that answer in my OP

Anonymous

@JasperLoy And then I begin to doubt my abilities,and then I get depressed :P

@Venus Oh, Didn't read.

@Venus You've posted that before my message right?

Anonymous

@Integrator Will I get your name by decoding?

2:50 PM

@Ashwin Yes, By any means are you from Pune?

Anonymous

@Integrator Bangalore

@Ashwin Oh, You're @Ashwin-gokhale !

@Ashwin I thought you're someone new!

Anonymous

@Integrator LOL no!!

@Ashwin Try your luck!

Anonymous

@Integrator Did you create the code?

2:52 PM

@Integrator Yep

@Venus Soon I'll post it, I need to some moderator (mostly robjohn) for that math110 and chinamath issue.

@Balarka Anyway, trying to use direct analogues of the fundamental group on spaxes with the Zariski topology seems wrong-headed. They're supposed to help understand the homotopy type of a space that's not entirely unlike Euclidean space; even Cech fundamental groups are born out of a desire to take this a little bit further to pathological spaces (that aren't entirely unlike Euclidean space). The Zariski topplogy is nothing like these.

@Venus I too think so because, in one meta post math110 said he is from china, though that proves almost nothing.

It doesn't help that the classical notion of homotopy is all wrong for Zariski topologized things, as they're all contractible, IIRC.

@Ashwin That's MD5 Hash for some string that includes my name.

@Ashwin It's highly unlikely that you could decode it.

2:55 PM

@MikeMiller I know I am really being vague about this, 'cause I haven't really done or thought anything serious about it. I felt that algebraic varities interact here, so I asked if there was such a notion.

Anonymous

@Integrator I am trying though

@Ashwin If you're able to do so, Then get ready to be popular.

Anonymous

:P

Don't take anything I am saying seriously. Realizing Gal(\bar Q/Q) geometrically is a dream of a child. I don't believe what I am doing would actually work.

If you're insistent on using algebraic topology for your dastardly non-Hausdorff purposes, you might look up Peter May's book on finite topological spaces. Maybe it'll give you the right ideas (and you can apply it to varieties over finite fields, say)

Anonymous

2:56 PM

@Integrator I am on a phone call wit Mr.Turing

@Ashwin Alan turing?

Anonymous

@Integrator He says only I can solve it :p

Anonymous

@Integrator One fascinating thing about Turing's is that Alan Turing's father had come to our remote village :D

@Ashwin Alan Turing died in 1954.

@Ashwin and you shaked hands with him?

Warning: I have no idea what's in it, other than that lots of spaces are weakly homotopy equivalent to a finite space, so that you can do lots of "classical homotopy theory" in this non-Hausdorff setting.

2:59 PM

@Ashwin and MD-5 was introduced in 1992

Anonymous

@Integrator But Turing was a Genius!