Mathematics - 2014-11-21 (page 3 of 3)

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5:00 PM

@r9m Absurd. Tunk-Fey gives derivations of the results.

Oh, when people dig and find out that people address me by Moron!

@r9m I was actually proposing that cleo not equal to Tuk-fey. (Chris's sis was the one who wants Cleo and Tunk-Fey to be equal)

@DanielFischer yes ! I think they are different too ..

Dinnertime, bbl.

@DanielFischer I've always wanted to ask you this question. What is your profession?

@HatMan she was claiming they were connected (outside Math.Se that is)

@DanielFischer enjoy dinner :-)

5:10 PM

@DanielFischer Am I right in thinking that $\pi_1$ of the suspension of a subspace $X$ of $\Bbb E^n$ can be computed without baby Van Kampen?

My idea is to identify geometric cones on $X$ with the natural cone $CX$.

@Balarka did you see @DanielFischer's answer to my question of ANT ? (main) :)

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integration calculus definite-integrals closed-form real-analysis improper-integrals polylogarithm

Ok, here's the rigorous proof : The suspension $SX$ with apex at $x_0$ and $x_1$ is homeomorphic to $\{x \in X, t \in [0, 1] : t \cdot x_0 + (1-t) \cdot x\} \cup \{x \in X, t \in [0, 1] : t \cdot x_1 + (1-t) \cdot x\}$

He is very good at integral problems. Is he really a high school student? If yes, I feel I am so stupid

@Venus he is GOOD !!

5:15 PM

@r9m link?

@BalarkaSen here

oh that yes i saw it and upvoted daniel

@r9m What make me amazed, he is a high student.

@BalarkaSen :) awesome answer !! :)

yes, nice, but i guess it can be shortened by some standard arguments

5:16 PM

This one too:

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integration calculus definite-integrals real-analysis trigonometry sequences-and-series improper-integrals

@Venus yes ! he was in this room this morning ! I was asking him how he got interested with these special integrals .. he replied he was motivated by the other answers in math.se !

Anastasiya & them are really amazing teenagers

yess !!

What, precisely, are you claiming to prove, @Balarka...?

@r9m Motivated by what?

5:18 PM

@Venus motivated by the other answers in math.se (evaluations of integrals)

@MikeMiller I am trying to find the fundamental group of the suspension of some connected space $X$. It's trivial, by baby Van Kampen but I am trying to do it otherwise.

It turns out we don't need baby Van Kampen when $X$ is a subspace of $\Bbb E^n$

$\Bbb R^n$, nobody writes E. And how so?

@r9m How old M.N.C.E. is? Can you guess?

@Venus I've no idea ! :o transcript from morning

@r9m How did you make a link like that?

5:23 PM

@Venus you mean how I fetched that link of transcript ? or how I formatted the link like that ?

@MikeMiller Let's see, I haven't formalized all of it. If $X \subseteq \Bbb R^n$ then $SX$ is homeomorphic to $\{t \in [0, 1], x \in X : x_0 t + x (1 - t)\} \cup \{t \in [0, 1], x \in X : x_1 t + x (1 - t)\}$.

How to format link like that? In the main page I use this [text](link), but it doesn't work here

Now WLOG let $\sigma$ be a loop based at $x_0$

@Venus [text](link) works here too ! I just did that !

@r9m Really? Last time I tried, it didn't work

SuperAbound

5:27 PM

@Venus works !

And take the part of the path that is not in the cone with $x_0$ apex, i.e., take $\sigma([0, 1]) - \{t \in [0, 1], x \in X : x_0t + x(1-t)\}$. Homotope this chap to $\{t \in [0, 1], x \in X : x_0 t + x(1-t)\} \cap \{t \in [0, 1], x \in X : x_1 t + x(1-t)\}$.

Geez! It apparently works

@r9m What make him/ her didn't want to reveal where he comes from? It makes me suspicious

And now take the whole loop (the tail homotoped to the boundary of the two cones) and homotope in to a point by straightline homotopy (the cone is star-convex, thus simply-connected)

Thus, every loop is contractible to the apex $x_0$.

Hence $\pi_1(SX, x_0)$ is trivial, thus $SX$ is simply connected.

Huh? Your second-to-last part doesn't make sense. You can't just homotope part of the loop.

@Venus there can be numerous reasons ! I don't know why they don't .. :)

5:31 PM

@MikeMiller erm. why not?

we can straightline homotope it.

Draw a picture. You're breaking the loop.

wait a sec i'm gonna post my drawing

@r9m His behavior looks familiar with the other user that I know here

Sensei @DanielFischer where are you from? Where did you attend your college?

user image

@MikeMiller that ^ is what i am thinking about

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calculus algebra-precalculus sequences-and-series complex-numbers integration limits derivatives

5:45 PM

And what do you do if your loop touches the other cone point, @Balarka?

@r9m That one too. Only 15 days here, he gets > 3000 reputations. Awesome user!

hello everyone!

@r9m Somehow this prof seems better than my idol Andre.

Brian M. Scott, Cleveland Heights, OH

251k 25 253 524

general-topology combinatorics real-analysis elementary-set-theory discrete-mathematics calculus sequences-and-series

@MikeMiller yikes. that's a problem.

Bounty lhf

0

Q: Sliding scale creation

This is possibly quite simple for a math-head, but I'm not one. :/ I'm trying to get to a formula that calculates a revenue share percentage based on turnover. My key values are as follows: $ Turnover % rev. share ------------------------------ 0 - 50000 30% 5001 - 10000 ...

algebra-precalculus

5:52 PM

@Balarka Because indeed if you can delete one point you get something homeomorphic to $CX$, which is contractible.

yeah. if you delete one of the apex it's homeo with CX. hmm. darn it.

thus i have only proved that a kicked CX is simply connected up until now.

6:05 PM

@Venus Andre's your idol ?! ;) okay :-) I like Brian Scott's answers too !! :D

i think i am convinced that it can't be done without baby V-K, @MikeMiller.

thanks for the checking.

I'm not convinced, @Balarka. Look up a proof that the sphere is simply connected and see if it gives you ideas.

the standard proof I know of again is baby V-K. Take $S^2 - p$ and $S^2 - q$ for distinct $p$ and $q$. These are cones, thus simply-connected. Intersection is a doubly punctured sphere which is connected. so $S^2$ is simply-connected.

but let's think of another proof.

The actually standard proof doesn't work that hard. Thwres a lemma that you can homotope to miss the North Pole. I'm suggesting you look at this lemma.

@Venus I often grin like my gravatar when I think about this :P ;) ..

6:12 PM

erm. stereographically through the north pole?

Huh?

what's the statement of the lemma? state it and i'll prove it.

Given a curve $S^1 \to S^2$, it is homotopic to one that's not surjective.

er, that'd mean it's nullhomotpic

Yes. That's the point.

6:16 PM

not sure how that'd help. it only shows that any loop on S^2 is contractible to some point. you need to fix a basepoint to work with the fundamental group, so it doesn't show, say, that if you have two loops \sigma_1 and \sigma_2 then they contract to some single same point.

@MikeMiller and why should i even believe it? take space-filling curves from [0, 1] \to S^2. surjective.

Yes. They're homotopic to something that's not surjective.

I didn't suggest you prove this. I suggested you look up a proof and generalize it to a proof that you can avoid a cone point of $\Sigma X$ for a connected $X$.

google's not turning up anything.

First result for "sphere is simply connected" is a PDF by John Lee. In it is a proof of the lemma.

@N3buchadnezzar what kind of $f$ are we investigating ? :-)

@MikeMiller page no?

6:29 PM

Dunno. I have faith in you.

i can't find it frown

@robjohn sensei any ideas regarding my question here ? (sorry for shameless advertisement)

Theorem 10.

@r9m Ya I like Andre. He helps me so much on probability questions and more important, he is patient to me. You wanna show off? Haha...

@r9m But I like that answer & I've upvoted long time ago

Nice answer!

@Venus show off ?! ;) not exactly .. Brian Scott left a comment ! .. I felt honored ! ;)

@Venus :-) thank you !

6:44 PM

@MikeMiller That's... essentially baby Van Kampen unfolded.

@r9m Ah, my bad for thinking negatively

@Venus they are two faces of the same coin ;) no worries .. I was aware of both sides when I tossed it ;)

That's fair, though the proof is shorter and more elegant because of the special case.

HelloMike

Why do you keep saying baby van kampen? What do you think is papa VK?

6:46 PM

@r9m But mine is Mobius-like coin :D

@Venus then we are not really tossing it are we ?! :P ..

@MikeMiller baby van kampen theorem : if $X = A \cup B$, $A \cap B$ path connected for $A$, $B$ open and simply connected, $\pi_1(X)$ is trivial.

Oh. Sure.

papa van kampen theorem : $X = A \cup B$, $A \cap B$ path connected, $A$, $B$ open then $\pi_1(X) = \pi_1(A) * \pi_1(B)/<\text{some blah}>$

@MikeMiller baby van kampen is much less pain in the neck than the papa version

and I thought babies were the pain in the neck !

7:01 PM

Wow. Internet shopping is fun.

One is more useful than the other and neither are too hard to apply. The proof is just much more painful.

@r9m I am proctoring an exam right now... I will try to look as I have time, but it may not be until tonight (here).

Fun stuff @robjohn... had to do that this morning

@robjohn okay ! thank you ! :-) ..

Hey people

@MikeMiller what does $\bra \ket$ mean in group theory?

7:20 PM

@UserX That looks like chemistry to me xD

7:45 PM

—Hello? Is this the anonymous FBI tip-line?
—Yes, Mr. Benson.

8:00 PM

Show that $\bra (1,2,3,4),(1,2)\ket=S_4$

$S_4$ is the full tetrahedral group but I can't understand what the LHS means

It's non-abelian and solvable

8:27 PM

Hi there!
Can anyone suggest me what journal I should submit my paper to?

2

arXiv

That's not a journal, and they told me to submit it to a journal, as it turns out to be not publishable, due to need of revision of my writing (not a matter of correctness)

@UserX

Is the group $\mathcal{G}$ with elements of the form $e^{\frac{2i\pi k}{n}}\, k\in\Bbb Z$ cyclic? How many elements does it have?

@VincenzoOliva who's they?

@UserX The arxiv moderation

8:46 PM

@VincenzoOliva so you'll never be able to publish it to arXiv?

@UserX They also said they will accept it if it is published in a journal.

9:19 PM

"We will not accept your article unless it's in a journal" is not the same as "You should submit your article to a journal".

Mike miller can you look on my questions above? Also, a cyclic group is the group containing elements that when raised to the minumum $n$ such that $x^n=1$. Is that $n$ the same as the $n$ in my elements of the group?

Can I abstractly use the $n$-th power?

I have no idea what you mean by the second thing. Do you have a book you're reading yet?

That's an exercise in MIT notes

And I started Artin but I haven't reached the part that talks about these yet

Can you tell me the specific wording?

Let $G$ be the subset of the complex numbers of the form $\exp (2\pi i k/n), k\in\Bbb Z$. Show that $G$ is a group under multiplication. How many elements does it have?

9:30 PM

So, yes, there are $n$ of those, you're right.

Is it a cyclic group?(that's my question, not the book's)

Can you think of something that should generate it?

$(z-1)^n$?

Committing to a challenge

@UserX Are you sure you don't want to do Math at uni :P?

What's $z$...?

9:35 PM

2πik for k integer is always 1. 1 raised to any integer power is 1. So the roots of the polynomial will be $(z-s^0)(z-s^1)(z-s^2)\dots(z-s^n)$ $n$ times

z is the variable of the polynomial that has the integer powers of the elements of G as roots

@Committingtoachallenge why do you ask that?

@MikeMiller does that make sense?

Committing to a challenge

@UserX Because you sure are doing a heap of it now

I bet that what I'm saying in the above message probably doesn't make sense :P

Committing to a challenge

You are talking 1.5x as much as I am on here haha, and I thought I talked a heap

It really doesn't...

@MikeMiller Can you answer it? Is G cyclic? If yes, what's it's name?

9:40 PM

@MikeMiller I asked confirmation that it is not a matter of correctness, and they said that.
Otherwise, they did say also: "Please submit instead to a conventional journal to receive the requisite feedback. " and " You should submit to a journal that best meets your needs and audience."

@VincenzoOliva have you asked for suggestions at mathoverflow?

@UserX Yes, it's cyclic of order $n$ (generated by $e^{2\pi i/n}$). It's called... the cyclic group of order $n$.

This is not a question that would be well-received on any SE site, @skull.

@skullpatrol Indeed I was not sure about that

perhaps even academia.SE

just to get some ideas :-)

I said any SE site specifically thinking of both of those. It's too localized of a question and would be downvoted and closed.

9:49 PM

he could still look around and see if anything similar has been asked

Can a non-mathematician(no degree) publish in journals?

@MikeMiller wow that was unexpected

Yes. Journals don't discrimate based on background (frequently things are reviewed blind - can't see name of author) but rather they discriminate on content.

@skullpatrol Perhaps, but I would have to find someone else claiming to have proved Robin's inequality, whom I'm afraid I would have already found

@VincenzoOliva try Annals of mathematics or Acta Mathematica

@MikeMiller What are your thoughts?

@UserX First the Annals, I guess?

@UserX Can I hope months won't pass?

10:00 PM

@VincenzoOliva No I don't think so. Annals is bi-monthly anyway

12 mins ago, by Vincenzo Oliva

@skullpatrol Perhaps, but I would have to find someone else claiming to have proved Robin's inequality, whom I'm afraid I would have already found

@VincenzoOliva What is this^ supposed to mean?

If he were to find such a someone he would have already found him by now

why?

He has this proof finished for about 7 months

@userx Well, nothing to do about it. Only one at a time,right?

For the record, my endorser's words on the moderation decision: "To me your article is it quite well written in latex, I have seen in arxiv much worse written papers with formulas in word, see e.g.

http://xxx.lanl.gov/pdf/math-ph/0404031

with many typos and unreadable formulas."

@skullpatrol What UserX said. Because I have done quite enogh research, so I doubt I would have missed it

*is quite

*enough

10:28 PM

@UserX Not that much, ahah. The main body was complete on September, but if we consider the addition of the lemmas and the improvement of the format, I've finished recently, since I did nothing on October.
Well, thanks for your help!

Going to bed now, bye all

later pal

10:39 PM

@Mike: Maybe things have changed since I last reviewed ... but, unlike the case in social science, in math journals I've never reviewed "blind."

Hello Professor @TedShifrin

Ok, I don't know. Regardless I think my claim on "content over author" is correct.

hi @skull

Odd how so many people here speak when they don't know of what they speak :D

:D

Certain authors carry a certain cachet, @Mike ... Let's just put it that way.

10:42 PM

I'd like a little more specificity. There are people publishing garbage who get it through because of the name?

I didn't say garbage ... But certainly older people with a name sail through grant approvals when younger people with no name don't ... and I've never liked that.

I believe that. But I (naively?) think the grant game is probably a bit different than trying to get a paper out...

not entirely, no

there are a ton of crummy papers published, that said, but rarely in the best journals

but there is an assumption that "high class" mathematicians with a rep should be approved and not so carefully scrutinized as someone "lower class"

Ok, fair.

"fair"

:-)

10:48 PM

You know what I meant.

11:03 PM

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11:19 PM

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