Mathematics - 2013-02-18 (page 1 of 3)

Can anyone help me showing this: Let $f:\mathbb R^n\to\mathbb R$. If for every closed set $G\subseteq\mathbb R$ $f^{-1}(G)$ is closed, then $f$ is continuous on $\mathbb R^n$. I think I should use that $f$ is continuous $\iff$ for every open set $S\subseteq \mathbb R$ $f^{-1}(S)$ is open, but I'm not sure how

Ethan

12:36 AM

$$\text{Li}_s(x)=\sum_{n=1}^\infty\frac{x^n}{n^s}$$$$-\frac{d}{ds}\text{Li}_s(x)=\sum_{p\text{ prime}}\ln(p)( \frac{ \text {Li}_s(x^p)}{p^s}+\frac{ \text {Li}_s(x^{p^2})}{p^{2s}}+\frac{ \text {Li}_s(x^{p^3})}{p^{3s}}+....)=\sum_{n=1}^\infty\frac{\Lambda(n)\text{Li}_s(x^n)}{n^s}$$

$$-\frac{1}{a}\sum_{n=1}^\infty\frac{\mu(n)}{n^s}\sum_{t=0}^{a-1}\frac{d}{ds}\text{Li}_s(e^{2\pi itn/a})e^{-2\pi i t b/a}=\sum_{n=0}^\infty \frac{\Lambda(an+b)}{(an+b)^s}=\sum_{p\equiv \text{ b mod a }}\frac{\ln(p)}{p^s}+O(1)$$

Peter Tamaroff

@MichaelCorleone Indeed.

$f^{-1}(C)$ beahves well under complementation, so I think you can prove it all by yourself.

Ethan

$$\sum_{p\equiv 2 \text{ mod 3}}\frac{\ln(p)p^s}{p^{2s}-1}=\frac{\zeta(s,\frac{2}{3})\zeta'(s,\frac{1}{3})-\zeta(s,\frac{1}{3})\zeta'(s,\frac{2}{3})}{\zeta(s,\frac{1}{3})^2-\zeta(s,\frac{2}{3})^2}$$

@draks...

Michael Corleone

@PeterTamaroff I'm sorry I'm not sure how though, a closed set is a set so that its complement is open, but not every open set is the complement of a closed set is it?

Peter Tamaroff

@MichaelCorleone $O$ is open if and only if $O^c$ is closed.

One usually defines a closed set as one whose complement is open.

Michael Corleone

Ok so that was my misunderstanding, so it is so that to every open set corresponds a closed set

Peter Tamaroff

12:52 AM

@MichaelCorleone Yes. You can define a topological space and in the same way you can define a "closure" space, and there is a one to one correspondence between "closure" spaces and topological spaces.

Michael Corleone

Ok thanks

Ethan

@draks...$$-\frac{1}{a}\int_{s}^\infty\sum_{n=1}^\infty\frac{\mu(n)}{n^s}\sum_{t=0}^{a-1}\frac{d}{ds}\text{Li}_s(e^{2\pi itn/a})e^{-2\pi i t b/a}\ ds=\sum_{p\equiv \text{ b mod a }}\frac{1}{p^s}+o(1)$$

$=\frac{\ln(\zeta(s)}{\phi_s(a)}, \text{ } \phi_s(a)=a^s\prod_{p\mid a}(1-\frac{1}{p^s})$

pourjour

plz some help

$\lim _{ x\rightarrow -\infty } 1-x-e^{ -2x }$

how can calculate the limt

Ethan

swap signs

$1+x+e^{2x}, as x->\infty$

pourjour

@Ethan is it by supposing $t=-x$ or what?

Ethan

1:06 AM

yes

user19161

The great Jonas is the Fourier God.

user19161

He doesn't hate you.

pourjour

@Ethan that's what I get $\lim _{ x\rightarrow -\infty } 1-x-e^{ -2x }=\lim_{\rightarrow +\infty}1+t-e^{2t}=1+\infty-\infty=?$

user19161

(removed)

Ethan

e^x grows much faster then x

pourjour

1:15 AM

@Ethan haha but I must change this idea into $x$ and $\lim$

Ethan

@pourjour What? I think just being familiar with some of these functions would help you.

pourjour

@Ethan I'm stuck bro :(

user19161

@ethan Have you sent those emails?

Ethan

$e^x=1+x+\frac{x^2}{2}..$

@JacobBlack No, I have been mostly buisy, what exactly should I write?

user19161

@Ethan Well, you should write what you think you should write.

user19161

1:21 AM

@Ethan Remember, focus on your school grades now and your SATs. Think of what college you wanna apply to. Plan for your future. And also deal with your other issues so they don't affect you.

pourjour

found it it was silly but it took me much time to realize the truth

user19161

@Charlie I have replied.

user19161

@amWhy I have replied.

amWhy

@JacobBlack I have replied, too! (twice in a day, OMG!!!)

500 more to go...then I'll slow down!

user19161

@amWhy That's what you said 10k ago.

peoplepower

1:33 AM

I find it odd that people set specific reputation goals.

user19161

@peoplepower We're all nuts here.

peoplepower

@JacobBlack After I got comments unlocked there is no desire for me to get rep.

amWhy

@JacobBlack I know...but I've got to focus on other things, and when I'm giving myself until 30 K to get back to business.

user19161

@ethan Feel free to share with me whatever stuff you want over email. =)

amWhy

@peoplepower comments unlocked?...it's just a number thing...I count endlessly, and make proportions, when I read (counting pages I've read, pages to read...), when I write, word counts...it's just a number thing...not really a rep-seeking thing.

@peoplepower What do you mean by getting comments unlocked?

user19161

1:38 AM

@amWhy Being able to comment everywhere.

peoplepower

@amWhy You know, enough rep to make comments.

amWhy

@peoplepower Dah...got it!

user19161

@peoplepower Too late, I answered first!

peoplepower

@amWhy I am just a simpleton; I always get surprised (in a negative sense :)) by the page number when I look up at it.

user19161

@peoplepower If you are simpleton then I am singleton, QED.

peoplepower

1:41 AM

@JacobBlack I'm empty. At least you have some substance.

amWhy

@JacobBlack replied (again!!!) :-)

Orange Harvester

so many emails without me!!! bawl

user19161

@amWhy I replied again 9000 years ago...

amWhy

I see that...I replied back!

Orange Harvester

bawl

amWhy

1:49 AM

@OrangeHarvester I'll get to you too! don't cry!!! I just don't write "text-y" style emails. I write more in conversation form. Promise, I'll connect. If not tonight, (for me that's in the next couple hours), then tomorrow morning!

Orange Harvester

@amWhy :-)

user19161

@OrangeHarvester Yes, I got email from 3 women today, and they all have a problem with me. =)

Orange Harvester

@JacobBlack GOOD!! (about the problem!) :P

user19161

@OrangeHarvester However, my only real problem now is A, B and C. QED.

peoplepower

@JacobBlack Are A,B,C entangled in any way?

Orange Harvester

1:54 AM

@peoplepower Yes.

user19161

@peoplepower A, B and C are my 3 secret wishes which change from time to time. So nobody really knows what they are except me.

Orange Harvester

@JacobBlack Blah. We all know all of them.

user19161

@OrangeHarvester Really? Then shh...

amWhy

@JacobBlack Are you counting me among them? (having a problem with you?)

user19161

@amWhy That is my secret.

user19161

1:58 AM

I enjoy saying secrets in non-secret ways.

amWhy

@JacobBlack Indeed. But in case...let me clarify: I don't have a problem with you. Perhaps you've gotten 9000 emails from women today, and 3 have problems with you :-D

user19161

@amWhy I also use various literary devices like hyperboles and parables, like hyperbolas and parabolas.

amWhy

@JacobBlack hehehehe...I do with language what I do with math...in many ways...

anon

2:22 AM

Let $f$ be a multiplicative function. If it exists, the asymptotic average is $${\bf E}(f):=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nf(k).$$ For a finite set of primes $S\subseteq \cal P$, where $\cal P$ is all the primes, let $\lfloor n\rfloor_S$ denote the number obtained by removing any primes $q\not\in S$ from the prime power factorization of $n$. Then heuristically,

$$\begin{array}{cl} {\bf E}(f) & =\lim_{S\to \cal P}{\bf E}(f\circ\lfloor\cdot\rfloor_S) \\ & = \lim_{S\to \cal P}\prod_{S}\left(1-\frac{1}{p}\right)\left(\sum_{n=0}^\infty\frac{f(p^n)}{p^n}\right) \\ & =\prod_{\cal P}\left(\sum_{n=0}^\infty f(p^n)\mu(p^n{\bf Z}_p^\times)\right) \\ & =\prod_{\cal P}\int_{{\bf Z}_p}f\big(|x|_p\big)dx. \end{array}$$ A curious formula.

Peter Tamaroff

3:16 AM

@anon Man. Can you help me? But be patient, please.

anon

I'll try.

Peter Tamaroff

Ok, please try.

Let $G$ have order $2k$, where $k$ is odd.

I must prove $G$ has a subgroup of index two.

Now, since $G$ has even order, $\exists g\in G\setminus \{e\}$ such that $g^2=1$:

Consdier the map $x\mapsto gx$.

We have that $x\mapsto gx\mapsto g^2x=x$

And since $g\neq 1$, $g\neq gx$, so that this never turns into a trivial cycle

Since no element is fixed, we get then a decomposition of this map into $|G|/2$ transpositions.

anon

you mean $g^2x=x$?

Peter Tamaroff

@anon Yes, corrected.

It follows that $g$ is not in the subgroup of elements $j\in G$ that induce even translations.

This means that this group, call it $F$, is proper.

anon

the subset of elements that induce odd translations/permutations will not be a subgroup

Peter Tamaroff

3:22 AM

And peoplepower told me something about the index being two since there are the same number of odd and even permutations, but I couldn't follow.

@anon I meant even.

Sorry.

Thus I want to show that $G=Fg\cup F$

Is that OK?

robjohn

@peoplepower I'll stop when I reach 20K!

Peter Tamaroff

@robjohn Hello, rob. Good to "see" you.

robjohn

@PeterTamaroff Hey there... been working on some questions

anon

there is a sequence $G\to S_G\to C_2$. the first map sends $a\in G$ to the permutataion $x\mapsto ax$. the second map sends a permutation to its sign. as your $g\in G$ has odd sign, this composition is surjective. If you write $F$ for the kernel of this composition (i.e. the subgroup of even translations), you know that $G/F\cong C_2$ by the first isomorphism theorem, and thus $[G:F]=|C_2|=2$.

Peter Tamaroff

@robjohn Mind to share?

@anon OK.

@anon It is surjective, yes.

Since the unit goes to $1$ already.

robjohn

3:26 AM

@PeterTamaroff I just finished answering this question. I may have misinterpreted the idea of simple. I wrote an answer with no prerequisites.

anon

"Image is isomorphic to Domain / Kernel" is one of those things you want to have engraved in your intuition in abstract algebra.

Peter Tamaroff

@anon Oh, I don't think I have gotten to the "first isomorphism theorem yet". Let me look at it. I'm sure I scanned through it.

It is not something complicated.

Blimey. But this exercise is even before the "homomorphism" chapter.

anon

Well, every $a\in G$ induces either an even or an odd permutation, every even permutation is in $F$, and if $x$ is any odd permutation then $gx$ is even and hence $x\in g^{-1}F=gF$. Thus $G=F\cup gF$ and $F\cap gF=\emptyset$. This is the barebones argument.

Peter Tamaroff

@anon Yes, I think that is what I wrote. Not so swiftly though! =)

Thank you, man.

@anon Are you in class "season"?

anon

wut

robjohn

3:32 AM

@PeterTamaroff I had never before seen a proof without calculus that $\lim\limits_{n\to\infty}H(2n)-H(n)=\log(2)$

Peter Tamaroff

@anon Well, my classes start next month

anon

Hard to do an uncalculus treatment of things involving logarithms and limits..

My classes started Jan 7th.

(We go by semester, not quarter.)

robjohn

@anon I think mine is okay.

Peter Tamaroff

@anon Oh, OK.

@robjohn I like your proofs.

As I was telling a friend.

You usually are the example of what I told him: "Don't use ready made recipes. Learn important yet simple things and use them wisely."

I'm thinking of using a picture of myself as a gravatar.

Opinions?

robjohn

@PeterTamaroff I'd scare people off if I did that

Peter Tamaroff

3:39 AM

@robjohn Ah, why? You have a friendly face. Much like @OldJohn

Hello everyone

Hello

@awllower Hey there

4:02 AM

Hi Guys!

How to make a (pie) sign in Chat jax ?

anon

chatjax uses latex as usual; you want dollar signs around \pi

devWaleed

ok

and how to write the equation in fraction form?

user19161

You should read a LaTeX guide.

John Smith

Can anyone help me with my graph question?

devWaleed

Tell me where is it?

user19161

4:04 AM

Read Oetiker's A Not So Short Introduction to LaTeX.

user19161

Just google it.

user19161

@anon looks as yellow as @robjohn.

devWaleed

but its a PDF file

robjohn

@JacobBlack I look yellow to you?

devWaleed

I was expecting a simple web-page...

user19161

4:07 AM

@robjohn Yes, maybe you need more Vitamin C.

2

user19161

@devWaleed Exactly, what did you expect?

robjohn

@JacobBlack Hmm... more carrots for Jacob

user19161

It tells you all you need to know about LaTeX for now.

devWaleed

$$ \frac13 2 $$

user19161

Well, I have told you the guide, the rest is up to you.

user19161

4:09 AM

There are no shortcuts in life.

devWaleed

see that? I am almost there tell me how to write 13/2

$$ \frac12 3 $$

please...

anon

\frac{a}{b}

devWaleed

curly brackets o.0

ok

$$ \frac{a}{b} $$

anon

http://en.wikipedia.org/wiki/Help:Displaying_a_formula
http://www.codecogs.com/latex/eqneditor.php

devWaleed

@anon -Thanks

robjohn

4:14 AM

@devWaleed $$\frac{\{a\}}{\{b\}}$$

anon

:\

user19161

@robjohn Haha, I was confused at first.

user19161

I don't like to install chatjax these days.

anon

these days? either it's a one-time-only thing (per browser) or you don't have your own computer, I would guess.

user19161

I install new OSes almost every other day for fun.

anon

4:17 AM

cool

user19161

Today Debian, tomorrow Ubuntu, next week Fedora...

user19161

Then the cycle repeats lol.

user19161

I may have done over 9000 installs by now lol.

user19161

Hey @math101 I counted that we have gone through at least 5 questions together.

math101

I think more than that

user19161

4:21 AM

They are not so trivial I would say.

user19161

I am surprised they gave you such difficult exercises.

math101

ik agree lol

user19161

Ramanujan is here, shh...

2

math101

And look who the wind blew in

devWaleed

@anon @robjohn $$ f = \frac{z^2e^42\pi^2m}{h^3} (\frac{1}{n1^2} - \frac{1}{n2^2}) $$ I was adding this to one of my question. ^_^

user19161

4:34 AM

Amazing that the wrong answer gets 3 votes.

user19161

Just a miscalculation though, not a big deal.

user19161

Ha, so I said "this is incorrect" and Don says "to claim this is incorrect without pointing out where is incorrect" haha.

user19161

I didn't even downvote the answer. It was so trivial I thought he would know. I think I should just keep my mouth shut.

Sanchez

link?

anon

http://math.stackexchange.com/a/306798/11763

user19161

4:43 AM

I have deleted my comment because it has served its purpose.

anon

how come your comment list on your profile is utterly empty, J?

user19161

Because I deleted all of them.

user19161

I just want to post good questions and answers.

user19161

After the comment has served its purpose it should be deleted.

user19161

There is no need to, but I think the site is littered with thousands of unconstructive comments.

user19161

4:45 AM

For example "+1 this answer is brilliant" is totally not constructive.

math101

hahahah it gives encouragement

user19161

In fact, often this comment appears on wrong answers and misleads voters.

Brian M. Scott

@JacobBlack I disagree completely. It says something useful to later readers of the answer.

@JacobBlack That’s a different issue altogether.

user19161

@BrianM.Scott Yes, but my obsession with minimalism takes over. =)

user19161

I am speaking for myself, you guys feel free to comment. =)

Brian M. Scott

4:47 AM

Ah. As long as it’s an expression of personal preference and practice, I’ve no problem with it at all.

user19161

Almost all statements are of such nature. =)

user19161

After all, is there an absolute right or wrong?

Brian M. Scott

@JacobBlack I don’t think so: I see an awful lot of statements attempting to prescribe others’s behaviour!

user19161

@BrianM.Scott OK OK, now you know. =)

user19161

Also, there is one type of comment I particularly hate these days...

user19161

4:51 AM

Comments like "I posted this answer first".

user19161

Or if there is a new user who asks a question in an inappropriate manner, rather than tell him how it should be phrased, they make some unconstructive sarcasm that doesn't teach him how to ask in future.

Brian M. Scott

@JacobBlack I dislike that one as well, though I don’t see it very often.

@JacobBlack And I despise that kind.

Ben W.

I really don't like the last kind either, makes the forum seem snobby.

A shame when the majority of the people here can be quite nice.

user19161

Of course, different people draw the line differently.

user19161

I certainly am not the type who thinks there should be absolutely zero sarcasm or jokes.

math101

4:55 AM

What bother me is that people are so bent on rules that they tend to be rude for some silly rule

I posted a question on MSE and cross validated SE

I didn't know that silly rule and they all started jumping on me as if I had just pulled the biggest prank

Sanchez

what rule is it?

user19161

Having said that, I am guilty myself of posting "+1 this answer is brilliant" comment myself too, sometimes.

math101

That you cant post the same question on any 2 MSE sites

Ben W.

At least not simultaneously.

math101

So if I posted the question on MSE then I cant post it on Cross Validated

user19161

4:58 AM

@anon Your current avatar really makes you godly lol

Sanchez

@math101, do you find the comments in that thread rude?

user19161

Where are the comments? I wanna see lol

math101

@Sanchez ok he/she edited his comment lol

Just realized that now

But either way you are right he/she didnt respond rudely

user19161

Ah, I don't like the adjective rude.

user19161

I don't use it to describe people usually, because I am not sure what it means.

math101

5:03 AM

It just felt overwhelming since everyone suddenly upvoted those comments

It means someone who is a jerk or a bitch

user19161

Ah, I ignore the votes on comments.

user19161

When I say I am not sure what it means, people might think I am autistic.

Ben W.

I interpreted it to be a philosophical question.

user19161

But I am not, I have pondered over the meaning so long that I see all sorts of problems using it, so I avoid it.

math101

Well rude is relative

Everyone has their own boundaries of what being rude means.

But its the same with many other words too

user19161

5:06 AM

I would rather describe exactly what the person did that made me upset, rather than attach these vague labels.

math101

lol you can do that if you would like

user19161

That way, we don't apply the wrong label to something that can be interpreted in over 9000 ways...

Ben W.

Does anyone here know umbral calculus?

Brian M. Scott

@JacobBlack That would be a pretty extreme reaction. I suppose that one might wonder about a mild form of Asperger’s syndrome, though even that’s pushing it.

user19161

@BenW. Sounds like the umbilical cord lol

math101

5:17 AM

Everyone has his/her issues

user19161

Yes, we're all nuts here, I am beginning to realize, lol.

math101

hahahaha My sis is trying to convince me that my dad is schizo, like some kind of mild form

My dad tries telling me that my uncle has bipolar

My friend tells me her dr diagnosed her with manic depression

user19161

You made 3 errors above.

user19161

schizo

user19161

bipolar

user19161

5:20 AM

manic

math101

My sister has OCD with regard to cleanliness

Everyone is crazy including me :)

user19161

You don't need to correct your errors in this chat lol

math101

hahahah I do

user19161

HAHAHA

Eric

5:38 AM

I just got chills reading the proof of Lagrange' Theorem for cosets

Mariano Suárez-Alvarez

you need to get out more

2

Eric

I mean wow, this is huge, and the proof is so simple

Actually i just got out earlier today.

user19161

@Eric Which result exactly?

Mariano Suárez-Alvarez

well

that is why I said more :-)

Eric

@MarianoSuárez-Alvarez ah, true

well it could also be because i am on my 5th cup of coffee

user19161

5:41 AM

One can think of many proofs while walking in parks.

user19161

I have never proven a difficult result sitting down.

Eric

@JacobBlack the one that says if $H \leq G$ then the number of left and right cosets if $H$ in $G$ is $|H|/|G|$.

user19161

Sitting down is for writing the proof after thinking of it. =)

user19161

Oh lying down helps too!

Eric

So does coffee! :)

user19161

6:14 AM

@Eric I prefer tea...

user19161

6:33 AM

Good night guys, enjoy your Sunday night sleep before the week begins...

The Substitute

6:48 AM

Hi.

What does it mean to say $a \asymp b$? What is the name of $\asymp$ ? How would you use this symbol?

Sanchez

their size are about the same

The Substitute

that can mean anything.

Matt

I have a quick question about the isomorphism theorems for rings. I have R a subring of S and J and ideal in S. then let I = the intersection of R and J. I want to show R/I -> S/J is injective. I've set up a homomorphism R -> S/J and shown the kernel is the preimage of J

So R/f^-1(J) is isomorphic to the image of R

Sanchez

7:07 AM

What is your difficulty?

Matt

if I is the intersection of R and J, is it necessarily true that I = f^-1(J)?

Sanchez

Yes

Matt

So then we know that the image of R is a subring of S so the map is injective