Mathematics - 2014-12-20 (page 3 of 8)

Venus

4:00 PM

@Chris'ssis Cauchy d'Alembert criterion, it's a new stuff for me. Thanks for sharing it. I'll learn it

Chris's sis

@r9m I'm a little girl too ... :-(

evinda

@DanielFischer That what I said above? We find the mid and then when we are for example at the first part of the array we look at one of the intervals $[0,mid-1]$ or $[mid+1,i]$ ? Or am I wrong?

r9m

@Chris'ssis NO ! =P You are demonic and monstrous Integral deity ! :P

Chris's sis

@r9m :-))))))))))) and much LOL

Daniel Fischer

@evinda The middle has nothing to do with it, We could have 100 negative elements and 9900 positive ones.

r9m

4:02 PM

@Chris'ssis LOL :P

Balarka Sen

no good question to answer today

meh

Daniel Fischer

@MikeMiller Don't know. It seems the OP doesn't see how to use the polarisation identity. On another note, I think your answer could use a couple of "for all".

evinda

@DanielFischer I meant the middle for each part... once for the part with the positive numbers and once for the part with the negative numbers... :/ How else could we do this?

Mike Miller

@DanielFischer Seems they got tripped up on something minor.

Daniel Fischer

@evinda Okay. It's not good to start in the middle of each part. Start at the beginning or end in each part (doesn't really matter which). You need to keep two indices (for each part), moving only in one direction.

@MikeMiller Yes, unfortunately we still don't know what.

Mike Miller

4:08 PM

Well, they upvoted and accepted, so I guess they're happy now.

Balarka Sen

$G$ be a group. When is it true that intersection of all normal subgroups of $G$ is trivial?

Daniel Rust

isn't the trivial group a normal subgroup?

evinda

@DanielFischer So do we have to check now each element of the first part with each other of the same part? And then the same for the second part?

Balarka Sen

eh, except the trivial group.

i'm confus.

voldemort

@LeGrandDODOM A werewolf hunter hat ;-)

Daniel Fischer

4:10 PM

@evinda No. That would be quadratic.

LeGrandDODOM

@voldemort what about the place in the background ?

Daniel Rust

@BalarkaSen groupprops.subwiki.org/wiki/Monolithic_group

skullpatrol

From what I understand @DanielFischer from this definition, zero cannot be written in scientific notation?

Balarka Sen

interesting

skullpatrol

"The special case of 0 does not have a unique representation in scientific notation" should read: "The special case of 0 does not have a ~~unique~~ representation in scientific notation" Am I correct @DanielFischer?

ccorn

4:17 PM

@skullpatrol If scientific notation is for inexact numbers, roughly indicating their accuracy, then zero would be outlawed right from the start...

Daniel Fischer

@skullpatrol If you take that description as a definition, then no - unless you say $\lfloor \log_{10} 0\rfloor = -\infty$ and write it as e.g. $1\times 0$ - but see "The special case of 0 does not have a unique representation in scientific notation, i.e., ..."

Balarka Sen

i'm trying to show that nonabelian group with smallest nontrivial subgroup has the property that every subgroup is normal @DanielRust

feels like it should be true

evinda

@DanielFischer So do you mean that we should have something like that?

i=0;
position=0;
while (i<m and B[i]<0) {
i++;
}
position=i;
if (y>0){
i=0;
j=1;
while (j<position){
if (A[i]<y*A[j]){
j++
}
}
}

But we have to increase also i, right? At which case do we have to do it?

Balarka Sen

@MikeMiller Covering spaces of $\Sigma_n$s for arbitrary $n$s seems to be pretty random

Mike Miller

What do you mean, @BalarkaSen? Which $\Sigma_n$s cover each other?

Daniel Rust

4:30 PM

13

Q: Groups with all subgroups normal

Is there any sort of classification of (say finite) groups with the property that every subgroup is normal? Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary. More generally, the group $\rlap{////////////////////////////////////////////////}\...

gr.group-theory

Balarka Sen

@MikeMiller I mean do we have an explicit classification of covering spaces of $\Sigma_n$s?

Daniel Fischer

@evinda Something vaguely like that. Not exactly like that but similar. You need to check whether A[position] == 0. And you have the wrong condition for incrementing j, think about that some more. It's probably better to first deal with only positive numbers.

Mike Miller

Of all covering spaces, or just compact ones? The former sounds hard; the latter is easy.

Balarka Sen

All

;)

Mike Miller

Sure. Every hyperbolic Riemann surface is the cover of some $\Sigma_n$ for $n>1$.

Balarka Sen

4:33 PM

What's your definition of hyperbolic Riemann surfaces?

Daniel Rust

What's $\Sigma_n$ here?

Mike Miller

The standard one, @BalarkaSen

@DanielRust Genus $n$ oriented surface.

Daniel Rust

Ah ok

yes, compact is easy

Balarka Sen

Compacts are easy.

Mike Miller

I would be surprised if you could write down anything nicer than what I said in the general case.

Balarka Sen

4:34 PM

They're just Galois.

@MikeMiller I dunno what a hyperbolic RS is.

Mike Miller

@BalarkaSen "They're just Galois?" That's not anything close to an 'explicit classification'.

Daniel Rust

@MikeMiller The quotient of the hyperbolic plane by a properly discontinuous action

Mike Miller

:P

Balarka Sen

@MikeMiller Any compact cover of $\Sigma_n$ is Galois, so the fundamental group injects onto the base space with normal image.

@DanielRust How is that even a Riemann surface?

Mike Miller

That's not an explicit classification.

You should clarify by isometries (or conformal maps, if you'd like), @DanielRust

Daniel Rust

4:38 PM

@MikeMiller true

and orientation preserving I guess

Mike Miller

I guess we should say conformal, because conformal includes that

Balarka Sen

@MikeMiller Well it's a step.

Mike Miller

@BalarkaSen For a compact topological space $X$, there is a covering map $X \to \Sigma_n$ iff $X \cong \Sigma_{1-k+kn}$ for some positive integer $k$. The problem with doing it in general is that there's no classification of hyperbolic RSes.

Balarka Sen

I'll just have to inspect normal subgroups of surface groups.

It's nothing sweating.

Mike Miller

lol ok

Daniel Rust

4:40 PM

just lol

Mike Miller

@DanielRust We can't both play bad cop.

Balarka Sen

OK, what is a hyperbolic Riemann surface?

voldemort

@LeGrandDODOM The Gaunt House ;-)

Mike Miller

@BalarkaSen A quotient of the open unit disc by a properly discontinuous action by conformal maps.

Daniel Rust

2

Q: Normal subgroups of fuchsian groups

This must have been known to the ancients, but I am having some trouble finding the references: what can be said (especially geometrically) about the normal closure of an element in a surface group? Especially an elliptic element (in an orbifold group)...

gr.group-theory gt.geometric-topology

Balarka Sen

4:41 PM

Why is that a Riemann surface?

Mike Miller

Why not?

Balarka Sen

OK, no, I am thinking of algebraic Riemann surfaces.

Daniel Rust

It's a complex curve (one complex coordinate)

Balarka Sen

Nvm.

OK, @MikeMiller. So, say, modular curves are hyperbolic Riemann surfaces?

Mike Miller

Yeah.

Daniel Rust

4:43 PM

@BalarkaSen did you see the question I linked? Fuchsian groups are a superclass of surface groups.

Balarka Sen

So the claim is that every covering space of $\Sigma_n$ is either $\Sigma_{blah}$ or a hyperbolic RS?

Mike Miller

Well, $\Sigma_{blah}$ is a hyperbolic RS.

Every covering space of $\Sigma_n$ $n>1$ is a hyperbolic RS. The point is that we can classify which ones actually cover $\Sigma_n$ when they're compact; not so easy in general.

Daniel Rust

$\Sigma_{blah}$ are precisely the compact hyperbolic RSs.

Balarka Sen

I don't see how. $\Sigma_1$ is just $\Bbb C$ quotiented out by {blah}. Not sure if it can be visualized as $\Bbb H$ quotiented out by something.

Mike Miller

It's not.

Chris's sis

4:46 PM

@Venus Can you take it from here?

0

A: Closed form of $\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$

Using series representation you specified above, I got that your integral gets reduced to $$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$ Can you take it from here?

Daniel Rust

$\Sigma_1$ is special, it's flat, not hyperbolic

Balarka Sen

Ah

Mike Miller

he should have said blah > 1 :)

Balarka Sen

$n > 1$

Hrm hrm hrm.

Mike Miller

blah > 1

Daniel Rust

4:46 PM

haha

Mike Miller

hehe

Daniel Rust

$n=0\implies\mbox{sphere}$, $n=1\implies\mbox{torus}$, $n\geq 2\implies\mbox{higher genus surfaces}$.

Balarka Sen

yes otherwise it doesn't make sense. $\Bbb R \times \Bbb R$ is not hyperbolic.

Venus

@Chris'ssis Maybe. Thanks. (+1) anyway :-)

Balarka Sen

although it's a covering space of $\Sigma_1$.

Daniel Rust

4:49 PM

the universal cover of a RS is either the sphere, the complex plane, or the hyperbolic plane, and that corresponds to the three regimes I outlined above

Balarka Sen

Cool beans. This looks like an interesting notion.

Chris's sis

@Venus I can give you the full support if needed. Try to finish it in your style. I'm curious how you do it. :-)

Mike Miller

Though @Daniel we're paying close attention to the metric/complex structure in this whole process. Maybe there's a more reasonable classification if we forget those?

Oh, nevermind, no.

Plane minus cantor set is hyperbolic, and if that's in there, like hell we're going to classify them.

Venus

@Chris'ssis OK

Daniel Rust

@MikeMiller oh sure, I just meant compact

Mike Miller

4:50 PM

well, what you said there is still true

I was just curious if we could topologically classify them

and I think the above says no

Balarka Sen

can fundamental groups of hyperbolic riemann surfaces be not hyperbolic?

Daniel Fischer

What is a hyperbolic group?

Mike Miller

I don't know what a hyperbolic group is, and before you start to define it, I don't want to hear it right now.

Daniel Rust

the fundamental group of the hyperbolic plane is trivial, is that a hyperbolic group?

Mike Miller

Ah, damn. The jig is up.

Balarka Sen

4:51 PM

i can define it.

$G$ be a finitely generated group.

Consider $Cay(G)$, a Cayley graph.

Equip it with word metric structure.

Now it's a geodesic metric space.

Daniel Fischer

No.

Balarka Sen

wat @DanielFischer

Daniel Fischer

That "No." was posted before Balarka's "Now it's ...".

Balarka Sen

a group is hyperbolic if every geodesic triangle in Cay(G) \delta-slim.

Daniel Fischer

@BalarkaSen That was my reply to "Equip ...".

Mike Miller

4:53 PM

The answer is automatically no, @BalarkaSen, because hyperbolic Riemann surfaces don't need to have finitely generated fundamental group.

Balarka Sen

@DanielFischer Well, just mark each edge with length 1.

@MikeMiller OK, hyperbolic riemann surfaces with finitely generated fundamental groups then.

Daniel Rust

ie the boring ones :P

(I kid)

Mike Miller

Well, @BalarkaSen, you just asked whether every finitely generated Fuchscian group is hyperbolic. You're the one who knows the definition of hyperbolic, so get crackin'.

Balarka Sen

oh noes

i can't do it.

Daniel Rust

I think you should extend the problem to looking at hyperbolic orbifolds :P

much more interesting

Balarka Sen

4:57 PM

@DanielRust do i look like someone who knows what an orbifold is?

:P

Mike Miller

No, or else he wouldn't ask!

Daniel Rust

@BalarkaSen an orbifold is a bit like a manifold, except you're allowed to make your charts out of discs modulo a finite group action.

(roughly)

so you can get marked points with an 'order' attached to them

or whole lines of marked points

Balarka Sen

blank stare

oh wait i think i know what you're saying

Mike Miller

I wouldn't advise explaining orbifolds, @DanielRust

He's just a kid.

Balarka Sen

@DanielRust consider the hyperbolic plane, and look at the action of PSL(2, Z) on it

Huy

5:00 PM

As one can see on the list of starred messages.

Daniel Rust

so take a disc and quotient it by an action of $\mathbb{Z}/n\mathbb{Z}$. As an orbifold, this space is different to the disc, even though they're homeomorphic

Balarka Sen

look at the two "kinks" on the edges of the fundamental domain of the action

hmm. i'm not entirely sure but i think the fundamental domain is a (2, 3, \infty) hyperbolic triangle

i will pummel whoever starred that

Daniel Rust

That's just a surface with cusps right?

so it's just not compact

Balarka Sen

@DanielRust oh ok

but i agree with @MikeMiller

you shouldn't give me rough ideas.

Daniel Rust

@BalarkaSen haha fair enough

Mike Miller

5:03 PM

Do you think about those guys much, @DanielRust?

Daniel Rust

@MikeMiller not too much. It was a problem posed at the end of a friend's thesis (for spaces which aren't manifolds)

so I've been thinking about it a little

Balarka Sen

@MikeMiller who says i am a kid?

Daniel Rust

but I think I'd need to learn about stacks before I could really dive into it

Balarka Sen

i'm 51, fyi

Mike Miller

I've never really had a reason to, so I only know the definition. But I guess they're indispensable in studying 3-manifolds.

Daniel Rust

5:05 PM

I turn 25 in four days

Huy

@BalarkaSen: 14 is more believable.

my covers are blown

Hey, all. ^_^

Hi pal ^_^

along with deck transformations...

Khallil Benyattou

5:06 PM

How goes it, @skull?

Daniel Rust

@KhallilBenyattou hey

skullpatrol

Fine thanks. How are you @KhallilBenyattou?

Khallil Benyattou

I'm doing good, @skull! I just learned about surjective, injective and bijective maps. Counting is a pretty useful thing to know of. ^_^

Hey, @DanielRust!

You got the Naruto hat, @Balarka!! :O

Balarka Sen

yes @KhallilBenyattou

Khallil Benyattou

How did you do that?!?

Balarka Sen

5:09 PM

the way it should be done

robjohn

@skullpatrol $0$ has a representation, just not a unique one.

Khallil Benyattou

Tell me! ^_^

Balarka Sen

answer a question, wait for it to be accepted, and wait for 12 hs

if no one upvotes, done

robjohn

@MikeMiller: finished with classes?

skullpatrol

I am not understanding that definition then @robjohn

evinda

5:14 PM

@DanielFischer I am confused now... I tried this:
`i=0;
position=0;
while (i<m and B[i]<0){
i++;
}
position=i;
if (y>0){
i=position;
while (A[position]==0){
i=position+1;
}
j=i+1;
while (j<m and A[i]<y*A[j]){
j++;
}
`

But it isn't right.... What could I change? :/

robjohn

@skullpatrol scientific notation is only unique for $x\ne0$

@skullpatrol what exponent of $10$ would you assign to $0$?

Daniel Fischer

@evinda Look at your test. Closely.

Chris Okyen

привет как дела

evinda

@DanielFischer Should it be A[j]<yA[i] instead of A[i]<yA[j] ?

skullpatrol

Daniel Fischer
8:17
@skullpatrol If you take that description as a definition, then no - unless you say $\lfloor \log_{10} 0\rfloor = -\infty$ and write it as e.g. $1\times 0$ - but see "The special case of 0 does not have a unique representation in scientific notation, i.e., ..." @robjohn

robjohn

5:18 PM

@skullpatrol If $-\infty\in\mathbb{Z}$ then you can do that...

Khallil Benyattou

Thanks, @Balarka!

Chris Okyen

So how can I figure out every combination that adds up to every number without repeats: 3 : 1+2, 1 + 1 + 1

robjohn

@skullpatrol I guess, with that as the definition you can't use $0$ as a mantissa... so you don't have a representation

Chris Okyen

summation or permutations

skullpatrol

Thank you @robjohn :-)

robjohn

5:21 PM

@skullpatrol I didn't say "overly" and I removed "pedantic" since it seems to raise hackles.

Chris Okyen

Combinations I think

Mike Miller

yup, @robjohn, I'm already back home... finished grading yesterday

err, two days ago

luckily my family lives in palm desert, so it's not as bad as it could have been. but with friday LA traffic near Christmas...

ccorn

Oh wow, I earned a spock's ears hat

evinda

@DanielFischer Or is there an other mistake?

Chris's sis

@Venus can you cope with the series?

Chris Okyen

5:26 PM

Anyone know how I can see all the values that summed up add to a value?

ccorn

@evinda Could you explain the problem again so I can offer help?

evinda

@ccorn I want to describe an algorithm that given an unsorted array $B$ that stores $m$ integers, and any integer number $y$, determines if there are two elements of the array of which the quotient is equal to $y$. The time complexity of the algorithm should be $O(m \log m)$.

ccorn

@evinda OK, thinking...

Chris's sis

@Venus Expand the terms of the series, and then multiply it by 2 to makes things easier. Then try to group the terms of the series in a clever way like $$\left(n^2 \log (n)-n^2 \log (n+1)+n \log (n)-n \log (n+1)\right)+\left(n^2 (-\log (n+1))+n^2 \log (n+2)+n \log (n+2)-n \log (n+1)\right)-\frac{1}{n}+1$$

Chris Okyen

...

skullpatrol

5:30 PM

They must have given you Spock's ears for your rejection of zero in science :P

Chris's sis

@Venus now, you can arrange terms from each bracket in such a way you get telescoping sums.

Khallil Benyattou

I've got a question about the integral you answered a few minutes ago, @Venus.
How did you guarantee that $\left| \cos \theta \right| = \cos \theta$ when factoring it out of the denominator in line $(2)$?

0

A: Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$

Let us evaluate the general form of the integral \begin{align} \int\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2-b^2}}&=\int\frac{a\sec^2t}{(a^2\tan^2t+a^2)\sqrt{a^2\tan^2t-b^2}}\mathrm dt\tag1\\[7pt] &=\frac{1}{a}\int\frac{\cos t}{\sqrt{a^2\sin^2t-b^2\cos^2t}}\mathrm dt\tag2\\[7pt] &=\frac{1}{a}\int\frac...

ccorn

@evinda positive integers?

evinda

@ccorn No, I assume that we can also have negative ones...

Venus

@N3buchadnezzar

He is level 5: Birch & Swinnerton-Dyer

I found it on AoPS

N3buchadnezzar

5:33 PM

@Venus I said he had 15k posts

Chris's sis

OUT FOR JOGGING

Venus

@N3buchadnezzar I can't see his profile. I must have an account for that

@Chris'ssis OK, OK

@KhallilBenyattou Good question. Andre has answered this kind of question in the comment (I can't locate it).

Maybe we just take the principal root?

Chris Okyen

I want to see all the possible sums to a number like for 3: 2 + 1, 1 + 1 + 1,

Venus

I think it is a common thing we're doing so

Mike Miller

Do you include 1+2 as different than 2+1, @ChrisOkyen?

Chris Okyen

5:38 PM

@MikeMiller Nope

Khallil Benyattou

Let me know if this seems legit, @Venus.
Is it because we're considering $x \in \left( -\infty, +\infty \right)$ (as the integral is indefinite) and with the substitution $x=\tan \theta$, we can deduce that $\tan \theta \in \left( -\infty, +\infty \right)$ and just take the interval $\theta \in \left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$ for convenience and over this interval, $| \cos \theta | = \cos \theta$?

Chris Okyen

So order doesn't matter?

if I only want 2+1 not 1+2 does that mean order doesn't matter?

ccorn

@evinda: Leaving some special cases aside: sort the array by absolute value. Copy the sorted array, with elements multiplied by $y$. Pass through both arrays like the comm utility, looking for a match.

skullpatrol

@ccorn They must have given you Spock's ears for your rejection of zero in science :P

Venus

@KhallilBenyattou I guess so, it seems that it makes sense

Daniel Fischer

5:40 PM

@evinda Yep.

Khallil Benyattou

Do you know of Andre's full username, @Venus?
(I might go and stalk some of his previous posts/comments to see if he talks about it there.)

Mike Miller

@ChrisOkyen You've described a very hard problem that's actually still being researched! The number of such decompositions of $n$ is the partition function $p(n)$. It's a very mysterious object!

The Wikipedia page I linked might have some stuff you'll find interesting.

Chris Okyen

for small numbers it may be simpler to write them out?

Khallil Benyattou

Has the layout of the Wikipedia page changed, @MikeMiller?
My one looks awfully different.

Mike Miller

Yeah, for small numbers you can pretty easily write out all the combinations.

@KhallilBenyattou Oops, that's rhe mobile site.

Khallil Benyattou

5:43 PM

Ah, got it! The mobile site looks pretty clean on a larger screen. ^_^

Mike Miller

@Chris It blows up pretty quickly though. $p(200)$ is something around 2 trillion, IIRC

evinda

@DanielFischer But don't we have to incerement also i? Because like that we just look at the first non-zero element with all the other elements... But we don't look for example at the second with the third one... Or am I wrong?

Chris Okyen

wouldn't all the combinations that add up to 9 be to many to write out

*too

Mike Miller

Oops - I linked the wrong page above

Partitions

Chris Okyen

1 + 8 2 + 7 ..... 5 +4 ..... 3 + 3 + 3 4 + 2 + 3 ....

Mike Miller

5:45 PM

$p(9)=30$, which you could write down by hand if you really wanted to. But it'd be tedious.

evinda