Mathematics - 2012-05-06 (page 4 of 5)

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6:02 PM

Since I'm done editing my answer, I should be going. See you guys later!

Byee!

Cya @JM

@JM Later! the teapot is gone :-(

6:19 PM

=(

Is there anyone who knows anything about elementary abstract algebra?

@N3buchadnezzar like what?

Some elementary things about normal subgroups

Give me a second to type in the question

" Let $H$ be a normal subgroup of $G$. Prove that $a^{(G:H)} \in H$ for all $a \in G$. "

x4d33746153706c306974

Why is relative speed important? can anyone please show me motivational problem?

@x4d33746153706c306974 What do you mean by relative speed? As in, one car is driving 60kph, another at 75kph, so the speed of the second relative to the first is 15kph?

@x4d33746153706c306974 If you are in a wild car race with the police, after robbing a liquor store. How fast the police car is travelling is not really interesting to you is it? What is relevant is the speed of the police car relative to your velocity.

6:33 PM

Good evening! Will the closed convex hull of compact in $C^{\infty}(\Omega)$ with the topology of uniform convergence be compact?

@N3buchadnezzar Unless you both are travelling at the speed of light in which case the chase will never end...

@N3buchadnezzar That's a good one. You've just committed a robbery + murder, and you decide to head for the border in your crappy '89 Nissan. You leave the store at 110 kph. Two minutes later, the cops leave the store at 140 kph after getting a tip about where you're headed. How long do you have to make peace with the fact that your life was meaningless and the world will be a better place with you gone?

4

@tb So my answer is useless?

@MattN No, I don't think so. There's just the fact that two definitions of the essential supremum norm are in common use. Let katari clarify.

@tb Thanks btw, the first line was not a typo.

6:41 PM

Good day Bill =)

@MattN With the definition I gave in the comments, note that the set $N_n = \{x \,:\,|f(x)| \geq \|f\|_{\infty} + 1/n\}$ is a null set, hence so is the union $N = \bigcup_n N_n$. We have $|f(x)| \leq \|f\|_\infty$ for all $x \notin N$, in other words $|f(x)| \leq \|f\|_\infty$ a.e.

: )

Jasper is dancing again?

"I think I'm brain dead or something, so I could use some help."

Then they need a shovel and a coffin.

2

Unfortunately, the gardening shop down the road only sells shovels, I checked.

@MattN When I become a teacher, and someone does bad on a test. I will give them their results, a shovel and a coffin.

Ah darn, just a shovel then and a origami coffin.

6:49 PM

@MattN It's hard to resist a request like that.

@BrianMScott Why?

@MattN The engaging way it's formulated.

Lol, you just set off my sark alarm : D

No, I'm serious.

I like the fact that she thought to ask herself about the converse, and I find the wording rather appealing.

But from a medical point of view it's too late once they're brain dead. : )

6:53 PM

@MattN A ton of movies, would argue against that.

Althought I doubt Zombies would be keen to learn Calculus.

At least Linda didn't choose the word as user name

We used to have an IDIOT and a Moron, but fortunately they re-considered...

Ayharabat ?

$\approx$, yes :)

@N3buchadnezzar Try Aryabhata. As I recall, it's the name of an early Indian mathematician.

I seem to have read quite a bit of his answers to tricky integrals, and everyone kept saying "Great job moron". Which I found a tad strange.

6:56 PM

@BrianMScott He's sometimes credited with introducing $0$, for example.

So I assumed it was his user name.

Fortuon Paendrag

@BrianMScott Thank you Brian, I was about to correct Nebuchadnessar :)

@N3buchadnezzar see also here

Fortuon Paendrag

On a tangent, It may be that the proof you gave me for the statement "A sequence of distinct rationals whose sum is finite has a subsequence summing to an irrational" contains a flaw. Could you help me rectify it? @BrianMScott

I can look.

Fortuon Paendrag

7:02 PM

0

A: field generated by a set

My proof is flawed. I will update it if I find a correct one. $S$ indeed generates $\mathbb{R}$. First we establish, with relative ease, that $\epsilon^2=\epsilon \implies\epsilon=0$ or $1$. Clearly, as $1 \in S$, it additively generates $\mathbb{Z}$ and before you know it, $\mathbb{Q} \in S$...

@robjohn I'm back. In fact I decided to post my answer to that question, which I think makes it much clearer. But the initial comment there (and other similar recent comments) have left a very bad taste in my mouth. Thankfully that has been temporarily alleviated by a night out with a much less abstract form of beauty.

@FortuonPaendrag Ah, that problem is much harder than the problem of just finding a subseries with irrational sum.

Fortuon Paendrag

Didnt we use the same technique though? "Pick an irational between the first term and the sum and add others(or not) as is wise?" @BrianMScott

@FortuonPaendrag Pretty much, yes. The difference is that you want to reach a specific target, and I just needed to reach any irrational.

If you need to hit a particular target, you have to have terms that let you do so.

@BillDubuque typo? "once could" = "one could"

@BillDubuque Indeed. One needs to take things on the internet with a grain of salt on the first square of the chessboard, and...

Fortuon Paendrag

7:09 PM

So, how would you begin the proof of my question to you? (Barring the Cardinality argument)@BrianMScott I apologize for not understanding faster.

@robjohn Thanks, fixed it.

@BillDubuque I saw, and upvoted :-)

@FortuonPaendrag Which question: the one about getting a subseries with irrational sum, or the one that you tried to answer on the main site?

Fortuon Paendrag

@Brian The former.

@robjohn Thanks, I was tempted to use Bezout for your amusement...

7:14 PM

@FortuonPaendrag Okay, you have distinct rationals $q_k$ such that $\sum_kq_k$ is finite. This immediately implies that a subsequence of the rationals converges to $0$, so to avoid ugly notation, let's just assume that $\langle q_k:k\in\Bbb N\rangle\to 0$.

@BillDubuque oh, ooh! what an opportunity. I should do that! ;-)

Fortuon Paendrag

Im with you.

(Hold on; I need to get more coffee.)

@robjohn Ha! I'll keep me eyes on whim.org for that.

Why is Wolfram so often misspelled as WoRlfram here?

7:20 PM

World-fram, clearly.

@TeddyBear I don't understand your comment: $f^\prime$ unbounded means $f$ cannot be Lipschitz. No?

I think I'm very hazy today.

@MattN true, but that's not what you say :)

@FortuonPaendrag Okay, I'm back. And on giving it further thought, I realize that it isn't quite so simple as I thought when I suggested the approach. I don't at the moment see a neat fix. I'll give it some more thought later.

@tb I thought I said $f^\prime$ unbounded hence $f$ not Lipschitz. But I need to re-read what I wrote. (I edited $f$ into the sentence you were criticising)

@robjohn Btw, I was quite pleased to see you run for mod. Of all my time on sci.math I don't recall many level-headed than you (even when confronted with strong Bezout ribbing!) I think you'd make a great moderator.

4

7:22 PM

@MattN I just saw. No complaints anymore :)

Fortuon Paendrag

@BrianMScott : Thank you Brian! I was furiously stomping up and down my apartment, trying to figure it out. It helps my mind that it is not a trivial problem.

Remind me next time you see me here, in case I forget.

@BillDubuque Thanks. I appreciate the support. I hope I'm not getting in over my head :-)

Fortuon Paendrag

Rest assured. I will! :)

@robjohn I'm sure you'd do fine. Just be careful that you don't let anyone pull your leg too much.

Fortuon Paendrag

7:25 PM

@MattN That picture of pooh running appears in my head whenever I see the initials t.b.! This is madness!

@FortuonPaendrag No. It's cuteness. : )

Fortuon Paendrag

How did you come up with this interpretation of t.b.s name?

Because, I sometimes see the name Theo used to refer to him(?)

Yeah, that's my real first name.

Fortuon Paendrag

Ah, that clarifies it! @tb

It is Tee-Bee.

7:29 PM

@FortuonPaendrag Srivatsan did. It fits the initials plus his real first name : )

@robjohn Oops, the above should say "I don't recall many more level-headed than you". And, in case you have doubts, I'm not pulling your leg when I say that! Such patience etc is key to successful moderation (alas, some mods have acted far too impulsively).

@BillDubuque heh, I was just going to ask about "past" mods :-)

@BillDubuque I agree 100%

@tb the rain is here

@leo unfortunately someone else sent that to you, we still have it :)

7:33 PM

@robjohn you here?

Resists. Temptation. To. Comment.

if yes, how did you parameterize the path $dz$ (back to the origin)?

here is about to start the rainy season :-|

i'm speaking of i.stack.imgur.com/mj8RA.png

@BrianMScott Do you happen to see a simple cardinal arithmetic argument why $\ell^1(\mathbb{R})$ has cardinality $\mathfrak{c}$? In a recent answer I used the fact that $\ell^1(\mathbb{R})$ embeds into $\ell^{\infty}(\mathbb{N})$ which I'm not really happy about because it's quite non-trivial: the usual proof uses the existence of an independent family of size $\mathfrak{c}$ on $\mathbb{N}$.

7:36 PM

@MattN Why resist?

I tried to count directly, but I failed.

@BillDubuque Because our friendly abuse regarding the weather is always the same. : )

@BillDubuque Oh, it was directed at me. We're almost neighbors and we have quite orthogonal opinions on what constitutes acceptable weather.

@tb Not immediately; let me think about it for a bit.

-2

Q: Measure theory problem.

Suppose $X, M, \mu$ is a measure space and $A_1, A_2, A_3, ...$ are sets in $ M$ such that each point of $X$ belongs to no more than d of the $A_k$ 's. then $\sum^{\infty}_{k=1} \mu(A_k) \le d \mu (\bigcup^{\infty}_{k=1} A_k)$ Plz help to prove it. It's hard to me

real-analysis measure-theory

:-(

7:37 PM

@tb Not enough fingers I suppose...

@tb try counting for a longer period of time

lmao

It's what he does when he can't sleep (instead of counting sheep hopping across a wooden fence...)

@tb neighbour. : )

tb integrates empirically

@tb Wait a minute. Can't you just look at $\prod_n[0,2^{-n}]$?

I feel violated.

Mariano Suárez-Alvarez

7:40 PM

After a while, one begins to enjoy innuendos...

@BrianMScott Sorry, I don't follow. What for?

@robjohn So, has MSE robbed your life of whim?

@BillDubuque No, I still have my whim :-) I refer to it once in a while

Wait a minute, not a removed message again. Pft.

@tb Every point in that product is a sequence in $\ell^1(\Bbb R)$, no?

7:42 PM

@Gigili they always make me feel as if I've missed something.

@robjohn Exactly so. I feel awful.

@MattN by a sheep?

@robjohn Usually it works the other way round, I understand.

Fortuon Paendrag

@robjohn @Gigili Ah, I merely thought what Brian had typed was a basis for $\cal{l}_1(\mathbb{R}$ . Later, I doubted myself.

@BrianMScott Have you ever played Settlers of Catan?

7:44 PM

@robjohn No, by someone's spelling of neighbours : )

@robjohn No, though I think that I may have heard of it.

@BrianMScott Sorry, how? $\ell^1(\mathbb{R}) = \{\langle x_t\rangle_{t \in \mathbb{R}}\,:\, \sum_{t\in \mathbb{R}} |x_t| \lt \infty\}$

Fortuon Paendrag

So you guys didnt miss anything important

@MattN but you've just misspelled neighbors :-p

@robjohn : P Now I have two people I can have endless arguments with : )

7:45 PM

:D

:D

@tb And that's true of any $x\in\prod_n[0,2^{-n}]$: $\sum_n|x_n|\le\sum_n 2^{-n}=2$.

sheep

@BrianMScott wood and sheep are two of the commodities that can be traded, and there have been several jokes made about it. They have actually appeared on The Big Bang Theory

@tb: Never mind.

@BrianMScott yes, but I want sequences indexed by $\mathbb{R}$, not $\mathbb{N}$.

7:46 PM

I misunderstood $\ell^1(\Bbb R)$, never having encountered it before.

Fortuon Paendrag

Ah, me too

Can anyone offer a clue about how to integrate $$\int_0^1 \frac{-Re^{\frac{i2\pi}{5}}}{(R(1-t)e^{\frac{i2\pi}{5}})^5+a^5}\ dt$$?

if you can read that

by parts, by parts :)

(don't take me seriously, Jeff)

@tb: But can't you just look at my sequences and fill in all the other terms as $0$?

Fortuon Paendrag

@t.b is it not true that all but countably many elements in a sequence of this space are zero ?

You beat me to it.

7:48 PM

@tb :D

Or for that matter look at characteristic functions of singletons?

@BrianMScott The trouble is the upper bound, not the lower bound. Every non-zero $\mathbb{R}$-vector space has cardinality $\mathfrak{c}$.

That's not a problem.

There are only $2^\omega$ countable subsets of $\Bbb R$ on which a point can live.

nevermind everyone... i forgot - i can't integrate that. I'm using it in a larger equation and it is a multiple the integral on the other side of the equal sign. :D so just forget i ever asked

And for each of those subsets, there are $2^\omega$ sequences in $\ell^1(\Bbb N)$. (Sorry to be so slow to realize what the actual problem was.)

7:50 PM

I'm surprised at how active chat is these days. I wonder why it took so long to get bootstrapped? I think the SE folks were surprised by that. It's not like mathematicians are less social than those on other SE sites, right? But seriously, does anyone have any clue why it took so long to ramp up?

@BrianMScott So you're saying that it boils down to $\#([\mathbb{R}]^{\leq \omega}) = 2^{\omega}$. I think I can agree with that.

That's indeed easier. Thanks!

No problem!

@BrianMScott But you might like the argument that shows that $\ell^{1}(\mathbb{R})$ embeds isometrically into $\ell^\infty(\mathbb{N})$, which took me about a week to find (and I found it a bit surprising).

Why we don't see Asaf here since some time ago

Mariano Suárez-Alvarez

Dear @skullpatrol and @BrianMScott if you are going to say that, say it... removing it after it gets read is, at least, unfair to Asaf

4

7:56 PM

Never mind.

@MarianoSuárezAlvarez You're right, but I didn't want to leave my comment hanging after skullpatrol deleted his.

.

don't have noticed they where talking about Asaf. Just asked by curiosity. Whatever...

@leo he decided to leave a while ago. He is pretty busy with his thesis at the moment.

@tb I see. Is he a Phd student?

7:59 PM

@tb Though he still patrols vigilantly for misapplied (set-theory) tags. :-)

Hello y'all.

=)

Hi, Peter.

@PeterTamaroff hi

Hullo, Peter.

@leo No, he's finishing his masters thesis this summer and he's going to start a Ph.D. thesis afterwards.

8:01 PM

@tb Good for him. :-)

@leo He left because someone flagged him. We don't know who because the coward didn't say that they flagged him. He's not going to come back in the near future. Just as well since then he'll have more time to study.

@tb He is still pretty active on the main site, however.

@MattN I see. Don't was me :-)

@BrianMScott and he may still patrol the chat transcript :-)

@leo Me neither : )

8:03 PM

Is he still the master of the room?

@mixedmath no, that's king robjohn now

@mixedmath no, he gave ownership to Gigili and she to me. Then she abdicated

all hail the King!

@mixedmath that's why robjohn's name appears in italics here.

oh - I never thought of that

how reasonable

8:05 PM

@tb Uh, never noticed.

@Gigili you've only noticed that the ground shakes a bit when I type? :-D

@Gigili Now you won't not notice anymore :)

And soon he'll be the uber-king by gaining mod-hood.

I'm not too familiar with how chat works compared to the main site, but I don't understand how one person could prevent it from prospering any more than one person could on the main site. Are there not safeguards here for preventing problems - just as on the main site?

@robjohn I get a chill when you type.

8:08 PM

@BillDubuque there's a flagging system and as soon as that's activated dozens of blue moderators come see what's happening. 10k+ users are also notified of flags which they can agree with or dismiss (there's a blue thing appearing instead of the orange one right now).

Asaf is good to the site. I mean he likes to discuss and to talk about math. And to debate. I think that we all have to discuss and defend arguments...

@BillDubuque the safeguards are called mods :-)

The owner has the ability to move comments to another room, or to connect a feed to a room, and to schedule events.

I can also unstar and pin comments

Have you ever googled "zerg rush"? Interesting.

@Gigili my browser melted!

@Gigili Haha, good one. Have you seen "askew"?

8:12 PM

@tb by the way, why some people put Ph.D. and others Dr.?

@robjohn I'm sorry.

Wooohoooooooo

@MattN Ah, great.

Someone just made my day!

what's the difference?

8:12 PM

No, my tomorrow. : D

Lecture has been cancelled.

@MattN My the day after today!

Which means I have a day off and all the time I want to do stuff.

@MattN Juhuu.

user image

@Gigili Yes : )

8:13 PM

@leo depends on the degree they got... I don't think there's a big difference except formalities and...the title, of course.

@robjohn Is this an Alien rip off?

Fortuon Paendrag

Surely you guys know about googling recursion then?

@MattN It's a Zerg and I've sent it after Gigili for melting my browser :-)

@robjohn I'll have nightmares.

@robjohn It is made in TikZ i think 8-)

8:14 PM

@robjohn : )

@robjohn Thank you, thank you.

@FortuonPaendrag Oh, interesting.

I didn't know that.

Fortuon Paendrag

Haha, I was quite amused when I found out

@Rob Can you lock a message so it can't be starred?

@tb I remember the history of a Russian professor. He gets his Ph.D. in France and his Dr. degree in Russia

@Gigili I first read that as so it can't be starved.

8:18 PM

@robjohn To me the square looks meaner than Zerg. I suppose I'll have to tone down my Bezout ribbing, eh? Not too mention all that "trash" talk!

I was thinking that Dr.$\gt $Ph.D.

The square is the meanest-looking critter I know.

@BillDubuque who took the trash out?

@BrianMScott That too.

@leo but the Russian doctor degree is much higher than an ordinary Ph.D. It corresponds more or less to what's a habilitation in old Europe.

8:19 PM

@robjohn No, who's on first.

@leo Not in the U.S. I believe that in Germany, among other places, there is such a distinction.

@Gigili I don't think that is possible, but I can keep unstarring it

It's possible. But I don't know by the room owner or mods.

@BrianMScott That was what I thought. A distinction

Can you make a dead user alive?

Haha?

8:21 PM

@leo In the US, Dr={PhD,MD} or Dr = MD U PhD, or something like that.

@robjohn but with PhD$\geq $MD

Not necessarily.

Psychiatrists have to have both

and What's the highest academic degree? at least in math

@BrianMScott In Germany and Austria a higher academic degree becomes part of the official name so you're basically forced to use it in official contexts. But you're allowed to substitute Dr. for Ph.D. (in Germany for sure).

The two are considered essentially equivalent.

Are you? I thought that I remembered reading in Der Spiegel that some visiting U.S. Ph.D.s had had problems with that.

8:25 PM

@BrianMScott Several friends of mine (who are German) have a Ph.D. and do use Dr. in Germany.

Okay. I wonder whether that's an even more recent change, or whether I'm completely misremembering.

Given that they like Papierkrieg a lot you probably have to ask to be allowed to do that.

Now that I can easily believe!

I imagine that's to do with the non-negligible amount of Germans who got their Ph.D. elsewhere but do not want to append Ph.D. to their names because it sounds a bit odd.

When I was in Germany I received several letters telling me that I have to use my title because I'm obliged by law to do so. I also had to fill out some forms to formally acknowledge that my degree is equivalent to a German Dr.

That would bother me a bit: to me it feels pretentious.

8:30 PM

I find it annoying, too, but it works wonders when dealing with banks and officials in Germany :)

I definitely avoid it when flying: I don't want anyone to mistake me for an MD in an emergency!

3

Or a dentist :D

I like The Chaz's comment here

@BrianMScott I wouldn't want that to happen to me either :)

Someone once asked my father (PhD in chemistry, college professor) whether he preferred Doctor or Professor; his response was something like 'Mister is always correct'.

8:35 PM

@AntonioVargas good one

I had two quite opposite grandfathers: one insisted on his Ph.D. which he got when he was around 50 with a major effort. I even have some letters of his ending with In Liebe, Dr. B.. The other grandfather got a Dr honoris causa and he sent back all the letters that used the title...

@leo I think that is a PhD

but it's the awards that count :-)

I'm off for a bit. Off to an art fair and then perhaps to a pre-birthday dinner. BBL

Tell me, who can I blame for the language "solve an integral"?

@robjohn enjoy :-)

cya @robjohn

8:40 PM

later, robjohn

nice

@robjohn How pre-? Mine's coming up in a week.

@robjohn Have fun!

@BrianMScott And you got your presents already weeks ago...

@MattN Very true! And fine presents, too.

Fortuon Paendrag

and these points count as early presents, surely. math.stackexchange.com/questions/141655/…

8:52 PM

@FortuonPaendrag I think that I actually got only 35 of them.

Fortuon Paendrag

Only? Thats one part in 20 of my reputation!

@BrianMScott sorry, I don't want to bore you with that, but here's the argument: given $n \in \omega$ and $A \subset \omega$ put $\varphi_n(A) = 1$ if $n \in A$ and $-1$ if $n \notin A$. This defines a linear functional on $\ell^1(2^\omega)$. Now if you identify $2^\omega$ with the set $S = [T]^{\lt \omega}$ of an independent family $T$ of subsets of $\omega$ the map $s \mapsto \langle \varphi_n(s)\rangle_{s \in S}$ gives an isometric embedding of $\ell^1(S) \to \ell^\infty(\omega)$.

(check isometry on the finitely supported sequences, which boils down to the very definition of independence)

@FortuonPaendrag Fair enough, but my mental scale is a little different: earlier in the week I came within 10 of getting 1000 in three days!

it would be great!

@tb Clever!

8:55 PM

@tb Hmm, that's what my advisor said to. If you append "prof. dr." to your name when writing to banks, institutions whatever it works wonders...

@BrianMScott well, after figuring this out, I found out that this is actually the origin of independence: Fichtenholz and Kantorovich used similar arguments to find the dimension of $L^1$-spaces which Hausdorff then streamlined by removing the functional analysis overhead and translating the constructions into purely set-theoretic language.

To the great benefit of many of us!

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