Mathematics - 2014-08-28 (page 3 of 4)

Mike Miller

5:01 PM

@Balarka I have a question for you.

Jasper Loy

@skullpatrol You must address me as Your Majesty.

lol

@MikeMiller askaway

drum roll...

and you don't have to be all sarcastic, @skull.

or i'll really consider ignoring you

skullpatrol

5:05 PM

@BalarkaSen i was kidding , i though we were pals again?

Balarka Sen

we are.

@Mike what is le question?

ah, he's gone.

skullpatrol

@BalarkaSen but "your highness" suits you, because it describes perfectly how you approach problems for a meta perspective...this is something to be admired in someone so young :-)

Mike Miller

@Balarka Show that there is a polynomial $p$ such that, for any finite abelian group $G$, $|\text{End}(G)| \leq p(|\text{Aut}(G)|)$.

skullpatrol

12 mins ago, by skullpatrol

drum roll...

Jasper Loy

@skullpatrol Do you like your current avatar? I don't like it, but that's just me.

skullpatrol

5:19 PM

I don't really look at it.

Jasper Loy

I much prefer the skull Charlie drew for you.

skullpatrol

icic

Balarka Sen

@MikeMiller That looks new. Don't really know how to approach it. Hmm.

Looks hard. Are you sure you are not giving me an open problem, @Mike? looks hard at him

Mike Miller

@Balarka It's open to me :) I'm sure it's known, though.

It suffices to prove it for p-groups, but this has proven tough.

Balarka Sen

It's probably waaay out of my league if it's open to you. But noted down at any case.

Will think about it.

@MikeMiller proving this for cyclic groups are enough, right?

Mike Miller

5:28 PM

@Balarka Unfortunately no. It's not hard in that case.

Balarka Sen

ah, sorry, sorry. you're right.

Mike Miller

I can so it for groups of the form $\Bbb Z_{p^n}^k$ by explicit computation.

But more general p-groups are proving devilish

Balarka Sen

explicit computation of auts of semidirect products are terribly hard, AFAIK.

Mike Miller

Abelian groups!

No semidirect products.

Balarka Sen

what?

Mike Miller

5:31 PM

I only care about this question for abelian groups.

Or rather I think it's intractable in general.

Balarka Sen

all righty. didn't notice that.

@MikeMiller does the extra structure of endos interact in that case?

Mike Miller

@Balarka My current plan of attack is to use the structure theorem for FGAGs. In general, I would need todo something like verify it for simple groups, then show it's stable under group extensions as well, which sounds hard.

If it's even still true.

Mike Miller

5:46 PM

@Balarka It is my belief that $p$ can be taken to be quintic; indeed I could take it to be $O(x^{4+\varepsilon})$ for any positive $\varepsilon$, I think. Quartics is not enough.

The current proof attempt, if it works, will give me an octic. But a polynomial bound is still cool.

Balarka Sen

Seems interesting.

Mike Miller

@Balarka Oh, oops, no, this problem should be doable with a quadratic. It should be doable linearly for p-groups.

@skullpatrol It's related to a recent MSE question

skullpatrol

icic

Balarka Sen

@MikeMiller have you played around with the idea that most simple groups have some kind of stabilizing Aut chain?

in particular, see if you can get some set-theoretic property of End(Aut(G))

Mike Miller

Nope. And the simple case is settled (recall, everything is abelian!)

Balarka Sen

5:52 PM

@MikeMiller Oh noes. I keep forgetting that.

I don't believe it's stable under extensions though. There is no particular property relating auts of a group to the auts of it's quotients.

Mike Miller

The reason this is likely intractable in general is that it will be very hard to preserve this under extension.

@Balarka Sure. When I say stable under xtension, I mean that if the two smaller groups are bounded by some nice polynomial, so is the bigger one. Set size is all that matters, not group structure.

Balarka Sen

the existence of $p$ is almost always guaranteed. my main interest is the bounding of the degree.

i don't see why it should be a quadratic, ps.

Mike Miller

@Balarka I can actually bound it by $O(x^{1+\varepsilon})$ in general now... assuming the bound is linear for p-groups b

@Balarka How is it "almost guaranteed"? :)

Ok, I gotta go now. Make sure to find a linear bound for p-groups by the time I get back, kthx.

mathh

6:51 PM

Hey everyone, why does Wolfram Alpha omit the solution x=-1 of the equation x^(x^5+37x^4+7x^3+84x^2+2x+9)=1 ?

Is it because there's a limit of solutions it can fit?

Khallil

Nice avatar, @mathh!

Balarka Sen

@mathh I guess it just took log of the equation in both sides and excluded the factor of $\log(x)$

Khallil

Hey, @BalarkaSen!

Balarka Sen

Hello

Khallil

Is there an algebraic way to solve @mathh's problem, @BalarkaSen? (I can only see approximations in the Wolfram result.)

Balarka Sen

7:01 PM

Ah, I get it @mathh

It took the usual branch of $\log$, i.e., $\log(-1) = i\pi$. In that case, when logged, it would never recover the solution $x = -1$.

@Khallil Just solve the quintic

Although, it's not solvable. So there is no algebraic solution, in one sense.

Khallil

I just realised that we have a fairly similar centre to our avatars, @BalarkaSen!

Balarka Sen

You're quite right!

Well, everyone seems to have the same center.

@Alizter

Alizter

@BalarkaSen Hi

Khallil

They do!

Hey, @Alizter!

Balarka Sen

@Alizter Hello

Khallil

7:05 PM

Good spot, @BalarkaSen!

Alizter

@BalarkaSen Hei!

Balarka Sen

@Khallil Let's do a research on the randomness of the SE-generated identicons.

@Khallil In fact, you're edges are the same as mine, just mirror symmetrized.

Khallil

Yes!

We are the complements of each other, @BalarkaSen!

(In terms of avatars, of course.)

Balarka Sen

Haha, yes!

The little squares in the center are filled in yours, while the edges instead are filled in mine.

The only nonsymmetric part is that your ninja squared are cut in half in my identicon.

Khallil

I see.

Balarka Sen

7:14 PM

*ninja stars

Khallil

Not to mention that if you rotate the edges of my own one, only then will our edges be the complements of each other.

Balarka Sen

@Khallil How's set theory?

Khallil

*Shuriken

I'm $\in \{$ Lazy people $\}$.

That's about as far as I've gotten.

All of my attention has honestly been consumed by Zelda.

I need to get back to it tomorrow.

Balarka Sen

@Khallil Let me see if I have a good counting problem for you.

Khallil

Coolio!

Balarka Sen

7:22 PM

Looks like I don't have any easy problem.

I was looking in here

Problem 41 on page 76 seems interesting.

Sawarnik

Oof, counting again.

Balarka Sen

@Sawarnik You should learn some.

@Khallil there's your counterexample.

@Swadhin's identicon doesn't have a prominent center, let alone a similar one.

Khallil

My counterexample, @BalarkaSen?

Balarka Sen

yes.

Swadhin

@BalarkaSen what's that supposed to mean?

Balarka Sen

7:27 PM

@Khallil there ^

21 mins ago, by Balarka Sen

@Khallil Let's do a research on the randomness of the SE-generated identicons.

Swadhin

@BalarkaSen oh.. :)

Balarka Sen

we conjectured that almost all identicons have the same center.

Swadhin

@BalarkaSen Can you please discuss the intrinsic and interesting properties of Random Walk?

Balarka Sen

@Sawarnik I read The Dancing Men. Excellent.

Sawarnik

:D

@BalarkaSen oh!

Balarka Sen

7:30 PM

@Swadhin I am little familiar with them. Not a probabilist. I have my own ways of thinking about random walks.

Swadhin

@BalarkaSen Then, I would like to know about them? Would you mind discussing?

Balarka Sen

I can only point you out connection of random walks with number theory.

Sawarnik

@Swadhin After a long time .. hello :)

mathh

I can see you're all very interested in the SE avatars, so take a look at this post (if you haven't seen it yet).

Swadhin

@Sawarnik Yes...

@Sawarnik Hello... :)

@BalarkaSen yes...that would be very much interesting

Balarka Sen

7:34 PM

@Swadhin OK. A generally known number theoretic function is the moebius function $\mu(n)$. $\mu(n)$ only takes the values $\{1, 0, -1\}$, the leftmost one if $n$ is square-free and has an even number of prime factors, the rightmost one if $n$ is square-free and has an odd number of prime factors, and $0$ if $n$ is not square-free at all.

It is conjectured that for large square-free integers $n$, $\mu(n)$ "almost" behaves like some random walk defined by associating heads to $1$ and tails to $-1$ in an unbiased coin.

Swadhin

I am having trouble in understanding, can you please bring in some interconnections?

Balarka Sen

There are empirically a good reason to think that. You can't usually compute $\mu(n)$ if you know $\mu(k)$ for $k \leq n$. The counterexample being that $\mu(pn) = -\mu(n)$ or $0$.

Mike Miller

@Balarka A friend and I achieved the bound. We'll write it up as an answer on the appropriate MSE question. I'll link you when it's up.

Balarka Sen

@Swadhin That is the interconnection.

@MikeMiller OK.

Swadhin

@BalarkaSen I am having trouble in understanding why the interconnection

Balarka Sen

7:38 PM

I am already interested.

@Swadhin Ah, you mean why would one care if $\mu(n)$ is random?

Swadhin

yes, I think so.

Balarka Sen

That's pretty advanced stuff. Ever heard of Riemann Hypothesis?

It is known that if $\mu$ acts like a random variable, then Riemann Hypothesis is true with probability $1$.

Swadhin

@BalarkaSen yes, I have actually

Balarka Sen

great.

Swadhin

@BalarkaSen tell me then.

Sawarnik

7:41 PM

@Swadhin Don't you think it is a bit too hard for our level? :|

Balarka Sen

The summatory moebius function, i.e., $\sum_{n \leq x} \mu(n)$ is called Merten's function and denoted by $M(x)$. It is known that if RH is true then $M(x) = O(x^{1/2+\epsilon})$ for all $\epsilon \geq 0$.

Swadhin

@Sawarnik do you think one should bind oneself by some level and not try to push it as hard as possible?

Sawarnik

Alright, ok :|

Swadhin

@Sawarnik why sorry? :)

Balarka Sen

However, by standard laws of probability, $\sum_{n \leq x} X_n$ is expected to be $\sqrt{x}$ where $X_n$ is the random variable associated to the coin (see above). This implies that $M(x)$ "should" be of order $\sqrt{x}$. That's all I can tell you with being as elementary as possible.

Swadhin

7:44 PM

why is this mathjax not redering?

Khallil

Ah, that counterexample!

Balarka Sen

@Swadhin here

@Khallil you're slow.

Khallil

Tell me something I don't know ...

Swadhin

I have used it, but I have to use it everytime some LATEX code comes up...

@BalarkaSen Okay, I have noted things down, I will study them, will ask you again, maybe on the intermediate level? I hope you won't mind sharing.

Balarka Sen

@Swadhin I won't mind.

Swadhin

7:46 PM

or rather enlightening this poor soul.

Balarka Sen

Or... I think you should leave it. It's too advanced at this level. Don't let them meddle your brain.

Digest what I said first.

Swadhin

I think, things are not as hard as you say, I checked out the blog you linked me to, I can understand the stuff, just having difficulty in understanding the notations, won't have much trouble in finding them over the net, would I?

@BalarkaSen

TRiG

5

Balarka Sen

@Swadhin It's not a blog, BTW. It's a forum.

Swadhin

@BalarkaSen yes, sorry for the mistake.

Balarka Sen

8:00 PM

@Swadhin You won't find everything on the net. I suggest you to read introductory books first.

fredsbend

@TRiG You got a spam flag for this??? I don't know the norms of this room so I can't really say. Just letting you know.

TRiG

Trying to work out what $128\sqrt{\mathrm{e}980}$ is anyway.

Swadhin

@BalarkaSen then, I will be asking that part to you. :) ASAT

fredsbend

I wouldn't even try. I have never understood logs.

skullpatrol

@TRiG 6606.4819

TRiG

8:01 PM

@fredsbend Did I? *shrug* I don't really know the norms of this room either. I pop in now and then.

Swadhin

@BalarkaSen suggest please. I do random reading, I know its harmful and impart incomplete knowledge, but i can't help it.

fredsbend

@TRiG It doesn't seem like you broke the rules:

51

Q: Main Chatroom Etiquette Rules

The previous owner of the Mathematics chat Asaf Karagila drafted some etiquette rules for the chat. They were deleted, so I am reposting them so that they might be observed. It's nice of you to drop by, after saying hello please spend a minute reading the transcript to see if there is an active...

Maybe bent the first one a bit.

TRiG

@skullpatrol I've only ever seen e used in logarithms and indices before. Actually, I have seen 2e, so it's not unknown to multiply by it. Rare, though.

@fredsbend Until such time as I'm banned, I'll not worry about occasional flags.

skullpatrol

Even getting banned is no big deal :-)

TRiG

I just happened to spot that image and thought people here might enjoy it. If anyone actually complains, I'll address it then.

fredsbend

8:04 PM

@TRiG lol. Yep, that's the way to do it. I'm not sure how many flags I've gotten, but I know its been at least ten in the last year.

@TRiG I liked it. I like those kinds of things. It makes chat rooms fun. Keep up the good work.

TRiG

@skullpatrol A mod once suspended me from chat for half an hour by mistake. I took it as a cue to get off the Internet and get some sleep.

fredsbend

I notice you have two stars for it.

Three.

skullpatrol

@TRiG These places are known as the black holes of productivity.

Swadhin

@BalarkaSen I am waiting for your suggestions, where do I start from?

fredsbend

@TRiG I'm not sure how many hours on average have to be logged to be banned my mistake.

Probably a lot.

@skullpatrol Tell me about it. I've been "working" for the last three hours now.

skullpatrol

8:08 PM

Abandon all hope ye who enter here!

Balarka Sen

@Swadhin sorry i was away for a while.

@Swadhin how much number theory have you studied?

Swadhin

@BalarkaSen no problem.

@BalarkaSen elementary...I am okay with the olympiad level

Balarka Sen

@Swadhin hmm. have you done modular arithmetic?

Swadhin

@BalarkaSen yes, I have.

Sawarnik

@swadhin have you given the rmo?

Swadhin

8:12 PM

@Sawarnik no, not yet, i think my knowledge is not complete enough yet.

Balarka Sen

@Swadhin can you prove that $p$ is a prime if and only if $(p - 1)! = -1 \bmod p$?

Sawarnik

@Swadhin No one's is ever complete. You should have given it for just a try :)

Swadhin

@BalarkaSen yes, actually I can.

Balarka Sen

@Swadhin show me.

Swadhin

@BalarkaSen don't you think its trivial?

Balarka Sen

8:14 PM

@Swadhin it somewhat isn't. show me your proof.

Swadhin

@BalarkaSen ok

if p is composite, then it is divisible by some other number q where 2<q<p-2

if (p-1)! was congruent to -1 mod p then it would also be congruent to -1 mod q, then (p-1)$\equiv 0$ mod q

is that enough?

you asked for proof of a theorem fondly known as Wilson's theorem

I have another one too.

the result is trivial for p=2

consider the case for p$\ge$3

let's consider the polynomial f(x)=(x-1)(x-2)...(x-p+1). f has degree p-1 and constant term (p-1)!

Its p-1 roots are 1,2,...,p-1

Balarka Sen

oof wait i got disconnected.

Swadhin

lets consider g(x)=$x^{p-1}-1$

ok....tell me when to resume

Balarka Sen

@Swadhin its not enough.

Swadhin

@BalarkaSen let me complete my second way. I will get back to the first one.

Balarka Sen

8:24 PM

ok.

the first one is not a proof, btw

Swadhin

g has also degree p-1 and leading term $x^{p-1}$

Balarka Sen

yes, i get it. you know the proofs well.

Swadhin

ok.

Balarka Sen

@Swadhin but i merely want to point out that the first one is not a proof.

you showed that if p is not prime then (p - 1)! is not - 1 modulo p.

which is not quite what i asked to prove =)

Swadhin

@BalarkaSen oh well, i get it, i prove something else. thanks for pointing out.

Balarka Sen

8:28 PM

you seem quite familiar with elementary number theory. do you have any background on calculus?

Swadhin

a little. I must not say I am strong, but I think I know the concepts, rigorous calculation in calculus is a bother for me.

Balarka Sen

@Swadhin do you understand Riemann Hypothesis?

Swadhin

@BalarkaSen a little, I was reading once, found it enticing, but didn't have the patience to read the whole of it, got scared by the demon calculations and symbols.

Balarka Sen

@Swadhin where were you reading it?

Swadhin

@BalarkaSen as usual, on the net

Balarka Sen

8:31 PM

@Swadhin OK, can you explain me the statement of RH?

Swadhin

@BalarkaSen very tough that...

Balarka Sen

So you mean you don't understand it?

Am I right?

Swadhin

@BalarkaSen not completely, I understand that the riemann zeta functions non trivial roots have real part 1/2

Balarka Sen

what is the riemann zeta function?

tell me.

Swadhin

the sum of the infinite series, $\dfrac{1}{n^s}$

but, I do not understand the complex plane, how do colours originate there?

Balarka Sen

8:35 PM

@Swadhin But I don't understand how you can claim that it has zeros with real part 1/2. It doesn't even converge for real part less than 1.

@Swadhin Colors?

Swadhin

@BalarkaSen that discussion is beyond the bound, I myself do not undestand it, I am telling what I read.

@BalarkaSen yes, I have often seen colouring of the graphs of complex plane, usually with VIBGYOR.

Balarka Sen

@Swadhin I am giving you bits to think about. Convince yourself that $$\sum_{n = 1}^\infty \frac1{n^s}$$ diverges for $\Re[s] \leq 1$.

If you are familiar with basics of calculus, you should be able to do that.

Swadhin

what is the last line with the symbol?

Balarka Sen

@Swadhin Those are just methods to plot it. Deeper colors are indication of the zeros and lighter colors are indication of poles. That's not mathematics, just a way of getting a 2-d immersion of the 3-d function.

@Swadhin $\Re[s]$ is the real part of $s$

Swadhin

@BalarkaSen yes, these are things that I do not get at all.

Balarka Sen

8:39 PM

@Swadhin don't let it bother you. it's not mathematics.

Swadhin

@BalarkaSen Oh, then I know it, I know the result.

Yes, it diverges, I understand very well that.

Balarka Sen

@Swadhin Knowing is not enough. Prove it rigorously and convince yourself.

Show me the proof as you understand it.

Swadhin

@BalarkaSen Cauchy Condensation?

Balarka Sen

@Swadhin Exactly.

Swadhin

and btw, $\sum_{i=1}^n \dfrac{1}{n^s}>1+\drac12+\dfrac13+...$, so its obvious isn't it?

Balarka Sen

8:41 PM

But how do you prove the divergence?

Swadhin

obviously s$\le$1

Balarka Sen

@Swadhin False. $1/n^s < 1/n$. Comparison doesn't help.

Swadhin

a fraction raised to the exponent of something less than 1 is not less than the fraction itself?

Balarka Sen

@Swadhin Ah, you mean $s \leq 1$.

Swadhin

yes, definitely, otherwise not.

Balarka Sen

8:43 PM

Then yes.

Swadhin

that is what you asked for didn't you

AndrewG

Herro. Got a group theory question. If $H \unlhd K \le G$, then $K \le N_{G}(H)$ but I don't think necessarily $K \unlhd N_{G}(H)$ -- does anyone know of a counter-example off the top of their head?

Balarka Sen

@Swadhin I asked for $\Re[s] \leq 1$, however.

Swadhin

@BalarkaSen now, how am I supposed to treat that? if complex numbers get involved in it?

Balarka Sen

That's an exercise for you to think about.

Swadhin

8:45 PM

actually, when complex numbers get involved, doesn't the numbers lose property of comparison.

Balarka Sen

@Swadhin BTW, are you familiar with any complex analysis?

Swadhin

@BalarkaSen I know complex numbers a bit, but what is complex analysis, I don't know.

Balarka Sen

@Swadhin Try reading Titchmarsh's Theory of Functions. You have to study complex analysis and number theory simultaneously if you want to understand RH.

AndrewG

@Pedro you're an algebra person, ya?

Swadhin

ok. does Titchmarsh serve the purpose?

Pedro Tamaroff

8:47 PM

@AndrewG Maybe...

Balarka Sen

Well, a more basic book of Spiegel-Lipshultz would also be good. More appropriate, actually, @Swadhin

AndrewG

Any idea on the above?

Swadhin

@BalarkaSen ok. let me see if i can download them

Balarka Sen

@Swadhin And try an elementary number theory textbook. Like Montogomery-Zuckerman-Niven.

Swadhin

@BalarkaSen ok.

Pedro Tamaroff

8:49 PM

@AndrewG What is above? I'm kinda lazy.

Balarka Sen

@Swadhin If you are not already familiar with this, here's an exercise for you : Prove that if $\frac{a^2 + b^2}{1 + ab}$ is an integer, then it's a perfect square.

Don't google.

AndrewG

Let $H \le G$ be normal in $K \le G$, then $K$ is in the normalizer of $H$ (right?) But I don't think K has to be normal in $N_{G}(H)$, just a subgroup of it.

Swadhin

@BalarkaSen I have tried this several times before, but everytime I start afresh, I get stuck.

Balarka Sen

it's kinda hard.

=D

Swadhin

Can you please provide me a memorable solution. I have solved it myself, but I forget it again.

Balarka Sen

8:52 PM

@Swadhin Rederive your solution and let me know. I won't give away this one.

Pedro Tamaroff

@AndrewG Let $H$ be normal in all of $G$. Then $N_G(H)=G$, and $H\lhd K$, but you can take $K$ not normal and be done.