Mathematics - 2014-05-17 (page 1 of 2)

user52932

00:02

Qs: if $\{p_n\}_{n=1}^\infty$ is a sequence of polynomials and $\sum p_n$ approaches $f$ uniformly on $\mathbb{R}$ as $n\to \infty$ then is $f$ a polynomial?

robjohn

@user52932 no. It is continuous, however.

unless the degree of the poiynomials is bounded, in which case, $f$ would be a polynomial.

r9m

@robjohn example of non polynomial $f$ ?

robjohn

00:19

@r9m actually, I was speaking in general. If the functions converge uniformly on all of $\mathbb{R}$, then eventually, they must all be the same polynomial modulo a constant function which decreases with $n$ (in which case, their degree is bounded).

r9m

@robjohn yas .. :)

got it ..

robjohn

@r9m on bounded domains, polynomials can converge uniformly to any continuous function.

r9m

@robjohn the stone weierstrass thm ?

robjohn

@r9m indeed

skullpatrol

@robjohn I'm glad to see you survived the great network crash of 2014 ;-)

robjohn

00:26

@skullpatrol It seems that I was out for longer than others. I could see answers appearing while I was still seeing a read-only main

skullpatrol

Me too.

user52932

01:02

@robjohn is there a simple counter example?

@robjohn the question says nothing about the degree of the polynomials

Ian

Hey @robjohn, are you busy now?

user52932

@robjohn or more specifically, lets say the polynomials do have a bounded degree then firstly how do we know that the partial sum of the polynomials must converge uniformly to a polynomial (if it uniformly converges)?

Mike Miller

01:33

hola @KarlKronenfeld

Karl Kronenfeld

@MikeMiller ey yo

Mike Miller

tensor is a stupid word

ಠ_ಠ

Does anyone know where I can find a proof that if random variables are independent, then the moment generating function of the sum is equal to the product of the moment generating functions?

Mike Miller

@KarlKronenfeld I'm annoyed at calling elements of certain tensor products "tensors"

it makes me want to punch a physicist, who are almost certainly to blame

Karl Kronenfeld

:)

The Substitute

01:36

off topic. hi. I'm trying to categorize all of the basic group constructions (such as subgroups, quotient groups, direct product). Perhaps it's debatable whether subgroups are a construction. I'm not looking for examples of groups (such as the integers with addition), but I'd like to know if I am missing any relatively basic constructions.

Mike Miller

semidirect product

free product. free product with amalgamation

those are probably the most generally useful

The Substitute

thanks Mike

Karl Kronenfeld

@TheSubstitute What sense of "categorize"?

The Substitute

@KarlKronenfeld I should have used the word "collect"

Karl Kronenfeld

aha

The Substitute

01:43

I wasn't sure if I was going to get called out on that :)

Mike Miller

there are some other important constructions but you won't see them for a while, probably

most constructions are just specializations to groups of things that work for lots of categories

Martin Sleziak

05:53

@MikeMiller Wikipedia mentions that: The free product is the coproduct in the category of groups.

And: The more general construction of free product with amalgamation is correspondingly a pushout in the same category.

But there is a warning in the same article: the free product is not the coproduct in the category of abelian groups.

Oh, sorry, I wasn't reading carefully enough. I thought you were asking about categorical versions of these constructions. While in fact another user was asking just which are most frequently used group-theoretic constructions.

Mike Miller

06:10

@MartinSleziak Ah, it's alright.

I actually wonderful, is the semidirect product also a general categorical construction? or some analogue?

Martin Sleziak

I do not know @MikeMiller. But the same question was certainly asked by others before.

Balarka Sen

I was serving through questions and I saw something like "Galois group of $\pi$ is $\Bbb Z$". The question is also cross-posted to MO and there is an article there, but it doesn't convince me (which I suspect is because of the informalness of the argument). Can anyone explain what's going on here?

It seems more doable to me what Arturo Magidin said, namely, that galois group of $\pi$ over $\Bbb Q$ is equivalent to galois group of a transcendental $x$ over $\Bbb Q$, which is the well know $\text{GL}_2(\Bbb Q)/\{\pm 1\}$

user116900

07:23

@ಠ_ಠ Are you here?

Parth Kohli

@JasperLoy Yes.

Sawarnik

07:42

@robjohn I have to prove that product of 5 consecutive integers can never be a perfect square. So, in those integers one factor would be 5, but only one, so its not a perfect square. Is this correct?

robjohn

@Sawarnik what if that multiple of 5 is 25?

Sawarnik

@robjohn Oh ok :(

robjohn

@Sawarnik I believe that Erdös proved that no product of 2 or more consecutive integers is a perfect square.

renyi.hu/~p_erdos/1939-03.pdf

karp2345

Hello everybody! Hi @robjohn @Sawarnik @ParthKohli . I am new to this chat, what do you do here?

@robjohn Ah sorry, my internet is damn bad.

So in the process I clicked send two times, and it posted two times :|

robjohn

@karp2345 aha :-)

karp2345

07:56

@robjohn Does your blue name means you are a mod? Then you can delete it :-)

robjohn

@karp2345 mainly math, but sometimes we are social (but this is a math chat, so the social part is often entertaining :-)

@karp2345 you can delete it, too.

karp2345

Ok :D

@robjohn Its 7 minutes now.

robjohn

@karp2345 hover over the comment and a triangle appears to the left. click on that

@karp2345 oh, too late for that one.

karp2345

:)

user116900

08:18

@karp2345 I knew someone by the name of Karp. Do you know me?

karp2345

@JasperLoy Nah :O

user116900

@karp2345 Phew!

user116900

@robjohn I just watched both The Amazing Spider-man 2 and Godzilla in the cinema, both are very good.

robjohn

@JasperLoy getting into the summer movie season

user116900

Yes, no need to delete, lol.

user116900

08:25

This room has the most deletions I believe.

user116900

That's because most of the folks here are crazy, lol.

user116900

08:37

I just answered a lhf, lol.

Parth Kohli

08:48

@JasperLoy Haha, nice answer.

Alyosha

@BalarkaSen Thanks.

Sawarnik

@JasperLoy And you get 25 points for that ... arghh.

I am tempted to downvote your answers. Its not fair :O

Balarka Sen

09:59

@Alyosha Uh, for what?

Alyosha

11:03

@BalarkaSen Answering my Galois-notation question.

Gabriel R.

Fun fact: $f \in C^\infty(\mathbb R,\mathbb R) \; \text{and} \;\forall n, \forall x \in \mathbb R, |f^{(n)}(x)| \leq 1 \; \text{and} \; f'(0)=1$ is enough to say $f=\sin$

Alyosha

That seems implausible.

Karl Kronenfeld

How about $f(x)=x$?

Alyosha

That's not bounded.

Karl Kronenfeld

oh, $n\ge 0$

Alyosha

11:17

Well, the Lagrange remainder of such a function tends to $0$.

Balarka Sen

@Alyosha Oh.

@Karl Did you see my question on galois theory posted in chat?

Karl Kronenfeld

you've posted a number over time

Gabriel R.

I'll translate the proof for you guys if I have some time

Balarka Sen

@KarlKronenfeld "this ain't my first rodeo"

5 hours ago, by Balarka Sen

I was serving through questions and I saw something like "Galois group of $\pi$ is $\Bbb Z$". The question is also cross-posted to MO and there is an article there, but it doesn't convince me (which I suspect is because of the informalness of the argument). Can anyone explain what's going on here?

►

Karl Kronenfeld

@BalarkaSen If the answer lies in what Qiaochu was suggesting, then it is beyond me at this point, though it is where I'd like head soon enough.

Balarka Sen

11:27

@KarlKronenfeld Yeah, I am not familiar with fundamental group vs. galois group duality either. (I was asking for a more understandable argument, actually)

It's Grothendiek's dessin de'nfant, isn't it?

Karl Kronenfeld

From what I gather, yes. Though, it is just a term to me.

Balarka Sen

@KarlKronenfeld Terms seems important while reading Grothedieck. I have heard from several people that loads similar things are out there like dessin de'nfant, anabelian geometry, grothendiek's galois theory which actually focuses on the same thing, but from different perspectives.

I am not sure which one actually works with Etale FG and GT duality.

Gabriel R.

math.stackexchange.com/questions/798939/… 2CHAINZ !

Alyosha

@GabrielR. Nice question.

Gabriel R.

Sawarnik

12:39

Nitish Kumar has resigned ... :( :'( ... sad day for Bihar.

skullpatrol

@Sawarnik thanks pal

ಠ_ಠ

13:22

@Jas yes

N3buchadnezzar

13:36

Mmmmm

$$
\int_0^1 \frac{\mathrm{d}x}{1+x^4} \sim \sin \left( \frac{\pi}{3} \right)
$$

Chris's sis

14:40

Greetings

$$\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(k+n+2)}(\psi(k+n+2)-\psi(n+1)) = \frac{ \pi^2}{6}$$

(newly created)

hmmm, did I miss the starting points for $k$ and $n$? It seems so.

It's this way $$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(k+n+2)}(\psi(k+n+2)-\psi(n+1)) = \frac{ \pi^2}{6}$$

Everything is just fine now!

user116900

16:13

@ಠ_ಠ Are your exams over?

ಠ_ಠ

@Jas No

@Jas I have 3 more next week

user116900

@ಠ_ಠ OK. Will you pass all of them?

ಠ_ಠ

@Jas no

Gabriel R.

how long is each one @ಠ_ಠ ?

user116900

@ಠ_ಠ The next name you choose, make it easy to type.

2

user116900

16:16

The three squares are all here, lol.

Alyosha

This is a long shot, but given any two elements $a,b\in G$, must there exist an automorphism such that $\text{aut}(a)=b$?

Balarka Sen

What's $\text{aut}(a)$?. @Alyosha?

ಠ_ಠ

@GabrielR. approx. 2 hours or so

Alyosha

The automorphism $\text{aut}$ applied to $a$.

Balarka Sen

@Alyosha Heh, I'd much prefer $\sigma(a)$

Alyosha

16:20

I guess not, but asked just in case.

Fine

Balarka Sen

@Alyosha Yes, this is nessesarily true.

Automorphism are all bijections.

Alyosha

Yes, but the set of automorphisms are a subset of bujections.

Balarka Sen

Crap, I misread the question.

So you are asking whether there exists a $\sigma$ such that $\sigma(a) = b$

Alyosha

Yes, for every pair $a,b$.

It seems very unlikely.

Balarka Sen

So you are asking for an automorphism that acts transitively on the group, right?

Alyosha

16:24

Yes.

Balarka Sen

Hmm, I can't give an off-the-top-of-my-head example, but I feel that this is false.

Alyosha

Yes.

Balarka Sen

@PedroTamaroff Give an example of a group $G$ with no element of $\text{Aut}(G)$ acting transitively on $G$.

Alyosha

A counterexample could probably be found by examining $\text{Aut}(G)$ for sufficiently small $|G|$, but I instantly forget most groups I learn of.

Balarka Sen

@Alyosha Same with me here.

Chris's sis

16:27

@BalarkaSen see my double series above :-)

Alyosha

There must be some interesting ones, but I've not found any yet (except those that are also fields given an extra operation perhaps).

Balarka Sen

@Alyosha Fields are easy enough to inspect, for $\text{Aut}(G)$ can be (sometimes) though of as the Galois group.

But I'd much like $G$ not a multiplicative/additive structure of any field $F$.

Alyosha

@BalarkaSen I'll wait for Pedro, then, though there is something on the mathworld page about Galois groups (mathworld.wolfram.com/GaloisGroup.html) I don't understand: why must $\sigma(x)=x$ for every $x$ in $K$?

Balarka Sen

It's defined that way, @Alyosha

Automorphisms of $L$ that fixes $K$.

Alyosha

Okay, thanks.

So $K$ is somewhat similar to the kernel?

In a group-actiony kind of way.

Balarka Sen

16:34

Well, kinda. Automorphisms restricted to $K$ are identities.

So they are kernels under inclusion homomorphisms, yes.

Alyosha

By automorphisms restricted to $K$ you mean $\text{Aut}(K/K)$?

Balarka Sen

No, I mean that $\sigma$ is a galois aut of $L$ if $\sigma$ restricted to $K$ is identity. (Note that $K$ is included in $L$)

@N3buchadnezzar Close, but not quite. Not even a coincidence, I'd guess.

By restriction, @Alyosha, I mean to restrict the domain of the automorphism into something smaller.

Does that make sense?

Balarka Sen

16:49

@Alyosha Try $V_4$, by the way.

Darn, it was easy enough to make examples.

Alyosha

@BalarkaSen Yes, it makes sense.

Balarka Sen

Wait a minute.

Alyosha

Sorry, what's the wordy name of $V_4$?

Balarka Sen

@Alyosha Klein-4 group.

Hey @Nick

Nick

@BalarkaSen: BalSensai, how are you :D

It's been a long time, what have you been upto?

Balarka Sen

16:54

learning stuffs.

Nick

Aren't we all :D

But you perhaps learn more than others :D

Balarka Sen

Not really.

I know much less than others.

Nick

I guess by the time you're my age, you'd have already conquered a great portion of mathematics

Balarka Sen

Nothing can be conquered in mathematics.

I'd, perhaps, have different perspectives of greater understanding. But it's unlikely for me to conquer anything.

Nick

:D Yeah whatever floats your boat. How's number thoery going?

Chris's sis

16:58

It looks nicer this way ... $$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(k+n+2)}\left(\frac{1}{n+1} + \frac{1}{ n +2 } + \cdots + \frac{1}{ n + k + 2} \right) = \zeta(2)$$

(in Ramanujan's style)

Balarka Sen

@Nick I have thrown myself out from number theory a bit. The thing I want to learn in future is equipped with much more tools than just NT.

Chris's sis

How many people said they want to become like Ramanujan? And then I wonder how many people said they want to go beyond what Ramanujan was ...

Nick

@BalarkaSen: Sounds good.

Balarka Sen

@Chris'ssis Many people went beyond Ramanujan, though, but never claimed they want to.

Chris's sis

@BalarkaSen Many? Which one?

Balarka Sen

17:02

@Chris'ssis Depends on what do you mean by "going beyond". By measure of the works, or the greater quality of the same branch Ramanujan worked in?

Chris's sis

@BalarkaSen by both the measure of the work and the great quality of his work ....

Balarka Sen

Sure. Alexander Grothendiek.

Chris's sis

@BalarkaSen He is a God for all ...

Balarka Sen

Also, Paul Erdos (classic)

@Chris'ssis Not really. One of the great ones, but not god.

Alyosha

Perhaps Neumann, though that's applied mathematics.

Balarka Sen

17:05

@Alyosha +1.

J. V. Neumann was polymathic!

Chris's sis

@BalarkaSen Keep in mind Ramanujan was not a human being ... :-)

Balarka Sen

@Chris'ssis Well, he kinda was.

=P

But, at that time, I think it was a lot to achieve. I mean, he developed the whole theory of algebraic number theory.

He is the father of modular forms, which were refined years after his original works, did you know that @Alyosha?

Alyosha

@BalarkaSen Yes, I found a paper explaining mock modular forms (which I didn't understand) a few days ago.

Balarka Sen

Ah, I heard about mock modular forms.

Alizter

@Chris'ssis How am I meant to do that integral you gave me?

Balarka Sen

17:11

5

Alizter

After much trial I looked at the answer and I have zero idea how to prove it.

Balarka Sen

This cat reminds me of something.

2

@Charlie?

Chris's sis

@Alizter Which one?

Alizter

$\int_0^1\frac{\log(1-x)\log(x+1)}{x(x+1)}\;\mathrm dx$ @Chris'ssis

Chris's sis

@Alizter This one requires some work though ... Did I give you this one?

Alizter

17:14

Yes :)

Chris's sis

@Alizter Oh, I see.

Alizter

@Chris'ssis Seeing as there are zeta functions in the answer. I didn't think I could do it.

Chris's sis

@Alizter I have a nice integral given on a Putnam contest I'm sure you'd easily do.

$$\int_1^{\infty} \frac{1}{e^{x+1}+e^{3-x}} \ dx$$

Alizter

@Chris'ssis Thank you will have a go now

Chris's sis

@Alizter OK

Pedro Tamaroff

17:27

@BalarkaSen That cannot work, since it is $\Bbb Z_2^2$, and ${\rm Aut}(\Bbb Z_2^2)={\rm GL}(2,\Bbb Z_2)$ will act transitively on it. In fact for any vector space $V$, ${\rm GL}(V)$ acts transitively on $V$.

Alizter

@Chris'ssis $\frac{\pi}{4e^2}$?

Pedro Tamaroff

Off the top of my head, I cannot think of one. But it suffices every automorphism has a fixed point.

Chris's sis

@Alizter Yeah :-)

Alizter

@Chris'ssis Thank you that one was lovely :)

Chris's sis

@Alizter :D

Pedro Tamaroff

17:29

Note that ${\rm GL}(2,2)$ has size $6$, and since it is nonabelian it is isomorphic to $S_3$.

@BalarkaSen I think I misread your question. You mean that no automorphism $t$ of $G$ is "cyclic" in the sense that picking $g\in G$ and looking at $1,t(g),t^2(g),\ldots$ gives $g$?

Alizter

@Chris'ssis Any more?

Alyosha

@PedroTamaroff Given $a,b \in G$, there exists at least $1$ automorphism of $G$, $\text{aut}(G) $, such that $\text{aut}(a)=b$. Give a counterexample.

Pedro Tamaroff

@Alyosha Ah, right. Then $V_4$ doesn't work.

Alyosha

How did you come up with that so quickly? Are there only one or two automorphisms of $V_4$?

Pedro Tamaroff

@Alyosha Six.

What...?

$V_4$ is a $\Bbb Z_2^2$.

That's a vector space.

Things are easy with vector spaces.

Alyosha

17:36

Could you explicitly sketch how you disproved the claim?

Chris's sis

@Alizter would you like a limit?

Pedro Tamaroff

I am bluffing though.

Alizter

@Chris'ssis Sure!

Alyosha

Okay. A vector space over which field?

Pedro Tamaroff

@Alyosha It acts transitively on the nonzero elements. =)

Alyosha

17:37

@PedroTamaroff Very nice.

Pedro Tamaroff

@Alyosha At any rate, the claim is false!

Chris's sis

@Alizter $$\lim_{n\to\infty} \left((n+1)^{(n+2)/(n+1)}-n^{(n+1)/n}\right)$$ This one is very beautiful. (no need for l'Hopital or Taylor series)

Pedro Tamaroff

For any automorphism $\psi$, $\psi(1)=1$.

So the action is never transitive.

Alyosha

Ah yes, thanks.

Pedro Tamaroff

Maybe you're looking for something else?

Alyosha

17:39

No, I'm not really.

I considered using that erroneous claim in a proof I was doing, but I found another way to do it.

Pedro Tamaroff

So @BalarkaSen maybe you meant something else, because for any group, the orbit of $1$ under ${\rm Aut}\, G$ is $\{1\}$.

@Alyosha What are yout rying to prove?

Alyosha

That $\text{Aut}(G)$ is cyclic $\Rightarrow$ $G$ is abelian.

It's simpler than I was making it.

Pedro Tamaroff

@Alyosha Oh. $G/Z(G)\simeq {\rm Inn}\; G$ is cyclic.

Alyosha

Consider the generating automorphism $\text{aut}_0(G)$, then there exists a $\text{aut}_1=\text{aut}_0^{-1}$, then do stuff and we have $(ab)^{-1}=a^{-1}b^{-1}$.

Pedro Tamaroff

@Alyosha Yes, or $G/Z(G)\simeq {\rm Inn}\,G\leqslant {\rm Aut}\, G$ is cyclic.

Alyosha

17:45

Very nice. I may have seen $G/Z(G) \approx \text{Inn}(G)$, but I wasn't familiar enough to think of using it.

Pedro Tamaroff

$x\in G\mapsto \{g\mapsto xgx^{-1}\}$. The kernel of this is $Z(G)$, the image is the interior automorphisms.

user116900

I see @pedro has become an expert in group theory, lol.

Pedro Tamaroff

@JasperLoy Hardly. I know a bit though.

Alizter

@Chris'ssis To my amazement, after some sloppy application of Stolz-C I get 1. I checked with M and I think that it's right :)

Chris's sis

@Alizter Yeah, that's the answer. :-)

Alizter

18:03

@Chris'ssis Anymore?

user116900

I keep mixing up alyosha, alizter and amr, lol.

Chris's sis

This one is marvellous (I hope you can do it) $$\lim_{n\to\infty} \frac {\displaystyle \cos 1 \cdot \arccos \frac{1}{n}+\cos\frac
{1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot
\arccos{1}}{n}$$

user116900

My life feels like a math problem whose solution I seem about to get but may never get. Wellness feels so near, yet so far.

user116900

I am lurking around the site, looking for low hanging fruits to answer, lol.

user116900

One of these days I must learn how to solve the cubic and the quartic.

Alizter

18:19

@JasperLoy The depressed cubic is very interesting.

Balarka Sen

@Alizter Much more interesting is $x^3 + c$

i.e., the binomial cubic.

Alizter

Yes factorisations of the form $x^3+a^3$ are interesting.

Balarka Sen

@Alizter No factorization. $x^3+ c$ is the general cubic and is as much as (more than, in fact) interesting as the general or depressed ones.

Hello, @TedShifrin

Ted Shifrin

Hi @Balarka @Alizter @Jasper

Alizter

Yhello

Pedro Tamaroff

18:25

@BalarkaSen I take you read my ping?

@TedShifrin Hey. I will have to go in some minutes, though.

Balarka Sen

@PedroTamaroff Yeah, I didn't realize that.

@PedroTamaroff I was reading them.

Ted Shifrin

Heya mr @Pedro

Gabriel R.

Salut @Ted

Ted Shifrin

Salut M @Gabriel ... Comment ça-va?

@Jasper: I'm not even getting thanks for my lovely mhf answers!

Gabriel R.

@Ted je suis très fatigué en ce moment (depuis la reprise des cours jeudi dernier). On prépare les oraux de concours (Polytechnique, ENS, Mines) et notre professeur nous a distribué un polycopié de 987 exercices (!). Et toi ?

Ted Shifrin

18:31

Il ne vous en a donné que 987? Pas beaucoup! :) bonne chance:)

N3buchadnezzar

oui oui oui

@TedShifrin medium fruit?

Ted Shifrin

Why not? @N3

N3buchadnezzar

Psychic floating fruit has never been my thing

Ted Shifrin

Very punny of you today, @N3:)

Alyosha

Is there a combinatorial way to prove $n^nm^m (n+m)! \le (n+m)^{n+m} n! m!$?

Stirling (inequality form) probably does it trivially, but that's less pretty than combinatorics.

Chris's sis

18:45

@Alizter are you done? (I was preparing to come up with some more questions to you)

Balarka Sen

Question : $A_5$ can be realized as Galois group of the covering $X(5) \rightarrow X(1)$. To what extent similar can be done, i.e., can all finite groups be thought of galois group of covering maps of Riemann surfaces?

Chris's sis

Keep in mind that one can be elementarily finished.

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$Mathematics$ Mathematics