Mathematics - 2014-02-12 (page 3 of 3)

Daniel Fischer

12:03

@PedroTamaroff Yes, but don't be too harsh, nobody can always remember to state all relevant constraints.

Matt N.

@PedroTamaroff I also bought Lee's Intro to topological manifolds but also have not even started reading. But I definitely count algebraic topology to one of my interests even though my knowledge of it is more or less nil.

Lol? Is Asaf reading the transcript and starring messages about him?

Pedro Tamaroff

@MattN. Dunno!

@MattN. Hm, that is a fancy topic.

Matt N.

Okay, I have to go and eat now. Byes for now!

Pedro Tamaroff

Cheers.

I will go out and jog.

Chris's sis

12:59

Greetings

Here is a very nice integral: evaluate by real methods $$\int_0^{\infty} \frac{t\arctan(t)}{\sinh^2(\pi t)} \ dt$$

Pedro Tamaroff

13:33

@DanielFischer

Daniel Fischer

@PedroTamaroff Anything up, or just saying hello?

Pedro Tamaroff

@DanielFischer I am continuing with the guide for showing $A_n$ is simple for $n\geqslant 5$ =P

Remember the last exercise was counting sizes of conjugacy classes.

Daniel Fischer

Okay.

Pedro Tamaroff

So now I have to show that if $H$ is normal in $A_5$, I can find $0\leqslant r,t\leqslant 1$ and $0\leqslant s\leqslant 2$ such that $|H|=1+20r+12s+15t$.

This follows by making $A_5$ act on $H$ by conjugation, yes?

Daniel Fischer

Yo.

And don't forget $\lvert H\rvert$ has to divide $60$.

Pedro Tamaroff

13:38

Aha.

The bounds on $r,s,t$ come from counting transpositions, $3$-cycles and $5$-cycles in $H$..

i.e. I have to prove $H$ has at most two noncommuting $5$ cycles, but all transpositions and $3$ cycles in $H$ commute.

Daniel Fischer

You could also start looking what divisors of $60$ you can write as $1 + 20r + 12s + 15t$ (with non-negative coefficients) at all. I'll be away for a while, don't expect speedy responses in the next time ;)

Pedro Tamaroff

Don't worry.

Alyosha

14:12

Does the abosolute value of the integral $\oint_C \frac{\cot(\pi z)}{z^2+b^2}dz \to 0$ as $S\to \infty$, where $C$ is the usual square contour centred at $z=0$ with side-length $S$ and $b \in \mathbb{C}$?

Pedro Tamaroff

Read this as cutest.

@TedShifrin !!

@Maik ?

Ted Shifrin

@Pedro! Welcome to my ice age!

Pedro Tamaroff

Birthday?

Birthday!

Ted Shifrin

Yes, but also "catastrophic" freezing rain/sleet/snow/ice ... all over the south of the US

Mike

dear god that's hideous

@TedShifrin Happy 53rd birthday!

Pedro Tamaroff

14:27

WAT.

Ted Shifrin

What are you doing awake at this hour, @Mike? 53rd? Where did you get that number?

Pedro Tamaroff

@TedShifrin 61st!

@TedShifrin I did some work on $A_5$.

Mike

@TedShifrin I just woke up. And I should have said 37th to make my point better :P

Ted Shifrin

Congrats, @Pedro. My personal favorite is to realize that $A_5$ is the symmetry group of the dodecahedron/icosahedron and then use symmetries to prove it easily.

Ah, yes, @Mike, that would have clarified things ...

Pedro Tamaroff

I proved the conjugacy class of three cycles have size 20, of 5 cycles have size 12, products of two disjoint transpositions have size 15 and with that prove it is simple.

@TedShifrin I have an icosahedron.

Ted Shifrin

14:29

Good ... As I say, I like to recognize that all from the three types of rotations of the dodecahedron.

Pedro Tamaroff

I now have to show that if $H$ is a normal subgroup, it has size $1+20r+12s+15t$ for some $r,t=0,1$ and $s=0,1,2$.

Ted Shifrin

Right, use your icosahedron. You should find obvious elements of order $2$, $3$, and $5$, and by the change of basis theorem, the elements of an order are conjugate to one another.

If $H$ is a normal subgroup and $a\in H$, then all the conjugates of $a$ must belong to $H$ !!

Pedro Tamaroff

I am doing this algebraically now!

Ted Shifrin

Pfeh @ algebraic.

Pedro Tamaroff

@TedShifrin Yes.

Ted Shifrin

14:30

Well, I could lose power for days (ugh), so you can have fun without me :P

Pedro Tamaroff

So I want to say the following.

Let $A_5$ act by conjugation on $H$.

Then 3-cycles are sent to 3-cycles, 5-cycles to 5-cycles, and products of two disjoint transpositions to products of two disjoint transpositions.

Mike

:(

Pedro Tamaroff

So I want to say $|A|=|1^{A_5}|+|T^{A_5}|+|5^{A_5}|+|3^{A_5}|$

Where $1$ is the identity, $T$ are transpositions, $5$ are five cycles and $3$ are three cycles.

Ted Shifrin

Well, you need to know that every element of $A_5$ has one of these orders.

It's not obvious that such would be closed under products, for example.

Pedro Tamaroff

Come again?

Orders?

Ted Shifrin

14:35

Why is a product of a transposition and an element of order $3$, e.g., already listed?

Pedro Tamaroff

Sorry, you lost me. Listed?

Ted Shifrin

Grr. When you list all the elements of order $1$, $2$, $3$, and $5$, why is this all of $A_5$? It's surprising ...

Pedro Tamaroff

Well, those are the basic even elts.

Mike

@PedroTamaroff I'm going to fight you.

Pedro Tamaroff

@Mike That'd be nice. Why?

Mike

14:38

"elt" :P

Ted Shifrin

LOL.

I prefer BLT.

Pedro Tamaroff

Bacon Lettuce Tomato?

Now I'm hungry.

Ted Shifrin

:)

Chris's sis

14:53

@robjohn you have to see this one $$\frac{1}{2}+\sum_{n=2}^{\infty} \frac{1}{2^{n+1}} \sum_{k=1}^{n-1} \binom{2^{n-k}+k}{k}^{-1} $$ It's too nice!

Mike

Update: don't try to work on your bed with your computer

Lest your room catch fire.

Ted Shifrin

Fire? Is the computer now smoking in bed?

Mike

It turns out mattresses aren't very good insulation @TedShifrin

(Though 'fire' is an exaggeration; it just turned off from being overheated)

Ted Shifrin

Oh ... I didn't realize you'd tucked the laptop into bed under the covers!

Mike

Hmm... I like that answer, but I feel it's the wrong one. Certainly whatever generalization shouldn't be contractible in general.

Ted Shifrin

15:04

Do you know about $K(\pi,n)$'s?

Mike

I know the definition, but virtually no properties.

But if a good answer would rely on that feel free to do so, it just means I have to work a little

Ted Shifrin

What do you have in mind? What would you do for $S^n$, $n\ge 2$? $\pi_n(S^n)\cong\Bbb Z$.

Well, you just need to be prepared for some totally abstruse constructions, I think. You want a $\Bbb Z$-space $E$ so that $E/\Bbb Z\cong S^n$.

So let's think about what the long exact sequence on homotopy would have to say. I've actually never thought about this question. And I am far from a topologist.

Mike

Hmm.

Danny

@robjohn robbie

@TedShifrin shiffy

Ted Shifrin

I think you won't like what you get. You'll have to have $\pi_k(E) \cong \pi_k(S^n)$ for all $k\ge 2$, which will make a total mess.

Hi @Danny.

Danny

15:12

@TedShifrin how is teaching?

Mike

Yes, I'm not convinced this is what we want. I'm having trouble articulating why.

Ted Shifrin

Don't ask, @Danny. We're in the middle of an ice catastrophe. 4 days of school canceled, almost surely, and I may not be able to leave me house for 4 days running.

Mike

What you describe is just a covering space of $S^n$, no?

Ted Shifrin

Not to mention the likelihood of losing power, they say, for up to days.

Mike

@Ted Wait, will that kill your hot water too?

Ted Shifrin

15:13

It sure looks like it, doesn't it, @Mike? Bad news.

Probably, @Mike. I have a gas water heater, but it may need an electric trigger to start.

The stove I can light with matches :P

Danny

@TedShifrin :/ oh, force majeure

Ted Shifrin

Definitely will kill heat ...

Mike

@Ted If the bad news is that it isn't, then that's good news. Since a covering space won't work at all - $S^n$ is its own universal cover!

Danny

@TedShifrin what state do u live in

Ted Shifrin

Well, if you want the space mod the $\Bbb Z$ action to recover $S^n$, doesn't it have to be a covering space? I suspect this is why no one talks about such things.

Georgia, @Danny, about 70 miles out of Atlanta.

Mike

15:15

@Ted Yes. And so I don't think a space such as you describe exists.

Danny

@TedShifrin ok...

Ted Shifrin

Well, @Mike, did I misread your question? Didn't you want $E/\Bbb Z \cong S^n$?

Mike

I think whatever it is must necessarily be more subtle, because covering spaces can't be the right tool.

@Ted No, no. Just some analogue of universal covers. I don't know what the analogue should be.

Ted Shifrin

So you don't want to mod out by a group action and recover the space ... So it probably is along Daniel's lines with some fibration. Still not convinced what the question really means.

Mike

I want something that I give a space, and it gives me an $n$-connected space that (probaly) fibers over $X$. And I want it to be universal in some category, and the $n=1$ case to reduce to covers.

If I get an appropriate analogue of the group action, all the better.

Ted Shifrin

15:21

So, if you're working with a CW-complex $X$, for example, you have to kill all its homotopy groups up to $n$. So this is already messy, but doable. Attach all sorts of cells. But I have no idea about the categorical status.

Mike

Now, taking universal covers is functorial, right?

$f: X \rightarrow Y$ induces $\tilde f: \tilde X \tilde \rightarrow \tilde Y$?

(I didn't want the arrow to be lonely and hatless.)

Ted Shifrin

Yeah. But what I'm saying now is a totally different sort of animal, I think. Just use homotopy lifting for your $\tilde f$, right?

Mike

That was my thought but I wasn't 100% on the details.

But if it's true I'll take that and move on.

If I can get something that satisfies my other requirements and then is functorial, I'd like that more than any analogue of the group action. But I can understand if I can't get functoriality.

Ted Shifrin

Well, $f$ lifts to $\hat f\colon X\to\tilde Y$ ($f_*(\pi_1(X))\subset \pi_*(\pi_1(Y^*))$), SO ...

I'm out of my league on this one, @Mike.

Mike

Hmm.

Let me edit the question for clarity and see if I end up grabbing one of the fancy-pants homotopy theorists.

Ted Shifrin

15:33

OK, I'm outta here for now.

Mike

Have a good day. Don't freeze!

Balarka Sen

Hey @Ted

@Nick What's up?

Did anyone say icosahedron? What about it? Invariants? Polynomials? Partition theory? Quintics?

Duality?

robjohn

15:54

@Chris'ssis that looks an awful lot like one of the expansions I use to compute $\gamma$.

Chris's sis

@robjohn hehe, true! :-)

robjohn

@Chris'ssis from this answer

Chris's sis

@robjohn Yeah. I saw that answer some time ago.

robjohn

@Chris'ssis yeah, but it looks like that double series

Chris's sis

@robjohn Yeap.

Mike

16:57

@PedroTamaroff My book isn't on campus D:

Chris's sis

17:35

@robjohn see $(29)$ here mathworld.wolfram.com/Euler-MascheroniConstant.html. We proved $(24)$ some time ago, and then it remains to connect $(24)$ to $(28)$.

robjohn

@Chris'ssis I am adjusting my answer to your question. I'll post here in a few minutes.

Chris's sis

@robjohn OK. Just let me know if you see an easy way to connect $(24)$ to $(28)$ (after finishing your work).

robjohn

@Chris'ssis I am not working from there...

Chris's sis

@robjohn ok

@robjohn Geez, I'm done! I missed something before!

@robjohn From a certain point I have 2 ways to get that result. I'm very curious about your way!:-)

@robjohn I think this is one of the most beautiful series I've played with that involves $\gamma$. The same thoughts for $(24)$ that is amazingly awesome.

I love this stuff!

Artem Kaznatcheev

17:55

Hi, are there any mods around?

Mike

@robjohn is a mod

Artem Kaznatcheev

@robjohn do you mind if I migrate this question from cstheory for you guys to merge with this question here

@Mike thanks

because it is completely off-topic for us, but it has an answer so I don't want to just close it. I would normally migrate to CS.SE, but it is a dup of a question here

robjohn

Using $(15)$ from http://math.stackexchange.com/a/344574 $$
\begin{align}
\gamma
&=\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{j\binom{2^k-1+j}{j}2^j}\\
&=\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{\binom{2^k-1+j}{j-1}2^{j+k}}\\
&=\sum_{k=1}^\infty\sum_{j=0}^\infty\frac1{\binom{2^k+j}{j}2^{j+k+1}}\\
&=\frac12+\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{\binom{2^k+j}{j}2^{j+k+1}}\\
&=\frac12+\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac1{\binom{2^{k-j}+j}{j}2^{k+1}}\\
&=\frac12+\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac1{\binom{2^{k-j}+j}{j}2^{k+1}}\\

@ArtemKaznatcheev I guess it looks okay

Artem Kaznatcheev

@robjohn Hokey dokey! Thanks.

Done

Chris's sis

@robjohn Very nice!

robjohn

18:10

@ArtemKaznatcheev I will merge.

@ArtemKaznatcheev The OP account did not get associated. I guess I need to talk to a comm mod

Artem Kaznatcheev

@robjohn it looks like the OP doesn't have an account here. He asked the original question that came to you on MO

robjohn

@ArtemKaznatcheev The account on the post you migrated looked blue, as if it were associated with an account here.

Mike

@robjohn I recognize the user.

Artem Kaznatcheev

@robjohn ahh, okay. No idea what's happening then!

Thanks for taking care of this! I'll chat with you guys more later

robjohn

@ArtemKaznatcheev I asked about it in TL

Pedro Tamaroff

18:57

Awesome

Mike

19:09

@Pedro I'm pretty sure the guy on the 1-0 law question is wrong about everything ever.

Pedro Tamaroff

@Mike Which guy? The one that answered?

Mike

Yah.

Pedro Tamaroff

I have corrected him twice already, I think.

Mike

He seems to think the language of groups includes the language of sets. But if that were true I'm pretty sure the problem would immediately fall.

Pedro Tamaroff

I have no idea about that, however. =D

Mike

19:12

Actually it's the other way around... you can define groups in the language of sets, and not the other way.

(I think.)

robjohn

@ArtemKaznatcheev Ah, I bet that his cstheory account is associated to his math account, but his MO account is not.

Artem Kaznatcheev

@robjohn that would make sense. Thanks for looking into it.

anon

@Mike For any given $n$, you can probably form $n$ $\exists$ and one $\not\exists$ and then form the appropriate equalities and inequalities to say that a group has order $n$. But that's for fixed $n$. I don't see how something like $\phi$ could be expressed in the language of groups, as it implicitly quantifies over an infinite number of potential cardinalities.

Artem Kaznatcheev

@robjohn I offer this question as a thank-you tribute.

robjohn

@ArtemKaznatcheev hopefully a comm team member can associate the math question with his cstheory account

Artem Kaznatcheev

19:24

@robjohn well, the user already got answers, so that's what matters. The account stuff is just for internet points, really.

robjohn

@ArtemKaznatcheev if there are comments or further answers, it would be nice if the OP were notified

Artem Kaznatcheev

@robjohn good point.

Mike

19:52

@anon right, but of course that's not what he's intending to do. you can't talk bijection a when your functions are homomorphisms.

Chris's sis

20:04

@robjohn pls don't forget to take a look at the link between the relation $(24)$ and $(28)$ above. Maybe you see a nice way to get from one side to another.:D

Paul Epstein

20:22

@Mike Have you read Khinchin's three pearls of number theory?

Mike

Only CF. What are the others?

Keep in mind, my taste in number theory is different than others.

Paul Epstein

@Mike, only CF -- is that a reply to me?

Mike

Yes - Continued Fractions.

@anon @Pedro [lol at troll answer](math.stackexchange.com/questions/662313/prove-that-x3-y3-z3-has-no-solut‌ions-as-simply-as-possible)

damnit

Balarka Sen

20:47

@Mike As I said, trollface

@Mike CFs can be hard to deal with, especially some which use elliptic modulars what-nots and algebraic number theory. I am not sure if Khinchin has them. Most works are done by Ramanujan.

robjohn

21:04

@Chris'ssis $$ \begin{align} \sum_{n=1}^\infty(-1)^n\frac{\lfloor\log_2(n)\rfloor}{n} &=\sum_{n=2}^\infty\frac{(-1)^n}{n} +\sum_{n=4}^\infty\frac{(-1)^n}{n} +\sum_{n=8}^\infty\frac{(-1)^n}{n} +\dots\\ &=\sum_{k=1}^\infty\sum_{n=2^k}^\infty\frac{(-1)^n}{n} \end{align} $$

Just count how many copies of the sum you need by looking at $\lfloor\log_2(n)\rfloor$

Chris's sis

@robjohn Very nice.

Charlie

Hi @chris

Chris's sis

@Charlie The great cat!!!! How are you doing? :-)

Charlie

@Chris'ssis couldn't be better, and you?

Chris's sis

@Charlie Playing here with some crazy stuff. :-)

Charlie

21:13

@Chris'ssis cool :)

Chris's sis

@Charlie I just found an integral where it seems Mathematica gives a wrong answer (even the newer versions). I was playing with that.

Charlie

@chris can't you correct it?

Chris's sis

@Charlie See this mathematica.stackexchange.com/questions/42195/…

@Charlie That one is a very ugly mistake.

Charlie

Oh

Chris's sis

@Charlie That integral you saw can be connected to another integral that has a special particular value that can be connected to the work of @robjohn above. It's an approach in Ramanujan's style. :-)

Charlie

21:25

Interesting :)

robjohn

@Chris'ssis what did I do now?

Charlie

Robjohning as always

Chris's sis

@robjohn You did nothing wrong. I was just sharing with Charlie some thoughts. :-)

Complex analysis

hi @robjohn

Hi @Chris'ssis

Chris's sis

@Complexanalysis Hi :-)

Complex analysis

21:31

wassup @Chris'ssis

Chris's sis

@Complexanalysis Not that bad. Playing with some stuff. How about you?

Complex analysis

@Chris'ssis attended 3 days seminar , got over today .

around 9 hrs each day

@Chris'ssis my head nearly exploded .

what are you playing with ?

Derek 朕會功夫

Hi guys

Quick question here,

Does anyone know how to start in this integration problem:

$$\int tan^{-2}x dx$$

Jessy Cat

21:53

It's the cot squared

Cotangent squared

Chris Okyen

Hey. earlier I asked a question about numbers in there number of combinations in termsd of solving using binary math / programming and was told to ask to go to SE main. If I ask this in Mathematic terms may I ask?

Jessy Cat

Then, you're just taking the integral of csc^{2}-1, which is easy peasy.

Derek 朕會功夫

@JessyCat How do you solve $\int csc^2 x dx$?

Jessy Cat

There's an anti derivative for it.

It's equal to -cot x

en.m.wikipedia.org/wiki/…

Chris Okyen

So I have my combinations here: pastebin.com/KDY64nG1
How can I determine with N digits how many combinations there will be?

Where no repetitons like 11 -> 11 or 123 -> 213 or 321 are allowed

The number of combinations goes for 2, 3 ,4 and 5 digits: 1 , 4 , 12, 24 combinations.. NOTE FOR THE FIVE DIGIT ONE IT SHOUD SAY 24 combinations.....

Hello @PaulEpstein

Paul Epstein

22:03

Hello @Chris Okyen

@Mike @Mike @Mike, failed number theory text alert mayday mayday.

Three pearls of number theory is a classic text by Khinchin.

That is the title.

So I didn't mean "three of Khinchin's number theory books."

Chris Okyen

Anyone able to help me?

Derek 朕會功夫

@JessyCat So you are saying I'll have to memorize it?

Chris Okyen

om going AFK but feel free to respond or Ping me!

Thank You So Much

user4140

@Derek朕會功夫 use quotient rule and the fact sin^2+cos^2=1

Derek 朕會功夫

22:26

@user4140 Quotient rule?

user4140

Quotient rule

In calculus, the quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist. If the function one wishes to differentiate, f(x), can be written as :f(x) = \frac{g(x)}{h(x)} and h(x)\not=0, then the rule states that the derivative of g(x)/h(x) is :f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}. More precisely, if all x in some open set containing the number a satisfy h(x)\not=0, and g'(a) and h'(a) both exist, then f'(a) exists as well and :f'(a)=\frac{h(a)g'(a) - h'(a)g(a)}{[h(a)]^2}. And this can be exte...

$-cot(x)=\frac{-(cos(x)}{sin(x)}$

Derek 朕會功夫

@user4140 Mine is about integrals...

Not derivatives

user4140

this is to see why the derivative of -cot is $csc^2$

see where I'm getting at?

Derek 朕會功夫

22:52

oh

wow I have to work backward to see it

this proves it but let's say I am taking a test or something like how would I know it's $-cot x$

Alizter

I had forgotten how fun it was to play with the mandelbrot set

user4140

@Derek朕會功夫 You should try plaung with cosines and sines... that's how I did it a week ago when my prof asked me.

Derek 朕會功夫

@user4140 plaung?

Kevin Driscoll

@Derek朕會功夫 He mean playing. 遊玩

user4140

@Derek朕會功夫 playing, sorry

Derek 朕會功夫

23:05

I see

Mike

@Paul Continued Fractions is probably better.

Derek 朕會功夫

@KevinDriscoll How do you know I know how to read Chinese

user4140

mabye he just translated it from google

Who is like the go to guy for category theory in mse?

Karl Kronenfeld

zhen lin is fairly active at MSE

also martin brandenburg

user4140

they both know their cat?

Kevin Driscoll

23:21

@Derek朕會功夫你的名字是中文的

Karl Kronenfeld

You can see the top answerers here. math.stackexchange.com/tags/category-theory/info

user4140

23:31

@KarlKronenfeld wow, that is sweet, didn't know that existed.

robjohn

@Complexanalysis hey there... sorry it took a while to get back. I was just wasting time on an answer to an ill-posed question.

Kevin Driscoll

23:48

@robjohn ill-posed actually or do you mean like its a formally ill-posed question?

robjohn

@KevinDriscoll they asked a loose question, but had something more specific in mind. Two people answered, but each answer was met with a reason that it would not work.

@KevinDriscoll It turns out there is no answer that can meet their needs.

Kevin Driscoll

@Robjohn Ah okay, that what I figured. Just wanted ot check you didn't mean something more formal like in the way that Fredholm equations of the 1st kind are ill-posed

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$Mathematics$ Mathematics