Mathematics - 2012-01-03 (page 3 of 3)

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6:09 PM

@tb: I have amended my Stirling answer. Let me know if that handles things better, or if there is something that is still unclear.

Looks like I got a couple downvotes recently.

Anybody know if someone's defined "noncommutative characters" that are $G\to R^\times$ instead of $G\to F^\times$ for $R$ a noncommutative division ring?

6:22 PM

How can I prove that $\aleph_0 \times 2^{\aleph_0} = 2^{\aleph_0}$ ?

I already know that $\aleph_0 \times \aleph_0 = \aleph_0$ and $2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0}$

remember the [0,1] bijection with [0,1]x[2,3]x...

that proves it doesn't it?

$2^{\aleph_0}\le \aleph_0 \cdot 2^{\aleph_0} \le 2^{\aleph_0}\cdot 2^{\aleph_0} = 2^{\aleph_0}$

Hmm, it wasn't [0,1]x[2,3]x .. it was union of those sets. But I see how this can be done: if we have f : R -> [0,1] then we can map (n, r) to 2n + f(r)

@Daniil I guess you're allowed to use Cantor-Bernstein.

yes sorry union

6:27 PM

@MartinSleziak Ah, true! Thank you.

[0,1]u[2,3]u... = Nx[0,1]

I totally forgot about that one.

I think hoovering and practicing the flute at the same time might be even worse than hoovering while listening to Brian Adams.

@MartinSleziak and for the same reason $\aleph_0^{\aleph_0} = 2^{\aleph_0}$: $2^{\aleph_0} \leq \aleph_0^{\aleph_0} \leq 2^{\aleph^{\aleph_0}_0} = 2^{2\times \aleph_0} = 2^{\aleph_0}$

Is this correct?

Thanks, QED.

@Matt I hate hoovering for that very reason.

You meant $2^{\aleph_0\cdot\aleph_0}$ there?

6:30 PM

@MartinSleziak yeah, sorry

@Matt and still worse would be hoovering, and practicing the flute to play Bryan Adams

When someone writes $a^{b^c}$ it is usually understood as $a^{(b^c)}$, not $(a^b)^c$.

@MartinSleziak do you, by any chance, know how to properly typeset that fancy `c' which denotes the power of the continuum?

You mean \mathfrak c?

$\mathfrak c$

Yeah, that one.

Thanks.

6:32 PM

or \mathfrack{c} to isolate the frackiness

@robjohn I'll take a look a bit later, thanks!

@AsafKaragila mathoverflow.net/questions/20882/… I thought you might like this link

@Daniil given that Asaf commented on the very answer you link to... :)

oh

damn, I did not notice that

6:48 PM

Nice, I'm in the acknowledgments of a paper 8-).

Next step: Be the author of one.

@JonasTeuwen congratulations

Well, it is not that good that it justifies congratulations... 8-).

What is the difference between a 2-tuple and an ordered pair?

none

that's the same thing

One of my students calls them duples.

7:03 PM

mathoverflow.net/questions/13031/how-to-define-tuples/13050

Strange question, IMO.

@CamMcLeman That's nice, I'll use that word from now on :)

@Daniil Thanks. As tb said, I have commented on that answer several times.

About the $\aleph_0\times2^{\aleph_0}$ question, there was a question about it on the main site some time ago.

@AsafKaragila This one seems to be pretty close: Prove that: $\aleph_0 \cdot \frak{c} = \frak{c} \cdot \frak{c}$

@MartinSleziak Cardinality of union of $\aleph_0$ disjoint sets of cardinality $\mathfrak c$ is even closer.

@Daniil but we proved that yesterday already, didn't we?

@tb Where and when?

Also, the meat came out magnificent. You should all hope that someday I'll be around to cook something fancy.

2

7:21 PM

2 days ago, by t.b.

So you have the union $\bigcup_{n=0}^\infty [2n,2n+1]$. Map the interval $(2n,2n+1)$ to $(1/(2n+2),1/(2n+1))$, map $0$ to $0$ and map $n$ to $1/n$ for $n \geq 1$. This gives a bijection to $[0,1]$

Ah.

True :)

@MartinSleziak What keywords did you use to find that question?

7:38 PM

@Daniil I tried "aleph_0" "2^{\aleph_0}" site:math.stackexchange.com and "aleph_0" "\mathfrak c" site:math.stackexchange.com

The links seem to broken - you should copy those phrases to google...

I see, thanks.

@MartinSleziak you'd need to replace ^ by %5E for the links to work

I have severe troubles writing $\aleph$ by hand.

(and probably { and } by %7B and %7C)

@robjohn This looks good to me (the calculation for $(7)$ is neat :). Here's one thing that might still be a point of contention: strictly speaking you don't show that you can swap the order of integration and summation, as you only estimate the contribution of the even terms. The odd terms could still produce a function that is so badly divergent that the rearrangement isn't justified...

@Daniil I don't :-P

7:54 PM

Israel is the perfect country for a set-theorist :D

@Matt 2x +1 from me today. Nice write-ups!

@tb Thank you!

x=0 ?

@Matt small fix to address Matt E's comment: You should consider $s|_{C_{\delta'} \cup O^c}$ which is continuous since it is continuous on $C_{\delta'}$ and constant zero on $O^c$. Now apply Tietze to obtain $s'$ continuous everywhere and bounded by $\|s\|_\infty$.

@tb Can I not argue that the support is a closed set contained in $O_\tau$?

8:03 PM

why 99% people answering the questions think abt the question a home work assignment and hesitate to provide the full answer

@coure2011, which?

@Matt You could, provided you use Tietze in the form: Suppose $s: C \to \mathbb{R}$ is continuous on the compact set $C$ and $O \supset C$ is open. Then there exists a continuous extension $s'$ of $s$ with compact support in $O$. The suggestion I made is a proof of this fact.

@Matt also look at the three times upvoted comment by commenter.

@Matt Did you see the point? You want to achieve two things 1) you want to extend $s$ to a continuous function $s'$ with compact support 2) you want the support to be contained in $O_\delta$ (so that you can control the integral $\int_{O_\delta \smallsetminus C_{\delta'}} |f - s'|$)

8:21 PM

@tb Yes. I had seen that point already but it didn't prevent me from making the same mistake I had made back then.

8:34 PM

@tb Oh. That's quite an easy fix.

@Matt Just to be on the safe side: Choose an open set with compact closure $O_{\delta}' \supset C_{\delta}$ before applying the fix (the $O_{\delta}$ you chose need not have compact closure)

So you have $C_{\delta} \subset O_{\delta}' \subset O_{\delta}$ and $O_{\delta}'$ has compact closure. Then everything works out fine.

Why can I choose an open set with compact closure?

Since $C_\delta$ is compact and contained in $O_{\delta}$ and since the space is locally compact: for each $c \in C_\delta$ there is an open set $U_c \ni c$ with compact closure contained in $O_{\delta}$ by local compactness. The $U_c$'s cover $C$. Take a finite sub-cover and put $O_{\delta}' = U_{c_1} \cup \cdots \cup U_{c_n}$. The closure of that set is the compact set $\overline{U_{c_1}} \cup \cdots \cup \overline{U_{c_n}}$.

Does $\mathrm{God}$ fit as a notation for a higher power?

8:52 PM

@tb I inserted the estimate (7) before removing the odd terms. They follow the same estimates as the even terms, so I modified (7), replacing $2k$ by $k$ without using cancellation

@AsafKaragila It is interesting that God is never mentioned as an exponential in the Bible.

@tb Thank you. Now I'm trying to figure out what will go wrong if the closure is not compact.

@robjohn When the Israelites fought some guys after their escape from the land of Goshen, they only won while Moses held his arms up. If that's not an exponential then I don't know what is exponential.

@tb Have you voted to delete on the "Do algebraists teach analysis sometimes?"

@AsafKaragila In Latin, exponere means "to put out". We had a lot of fun with that in high school.

@robjohn Hah!

@Matt Oh, that's my mistake: you already have the good form of Tietze given in your question, so no need for that move!

8:59 PM

@AsafKaragila It really means "to explain", but "ex" = "out" and "ponere" = "to put" so we sort of reassembled it :-)

@tb And the bad form is this?

@Matt No :) Here's the really good form (and the one you want to have). Let $X$ be locally compact, let $K \subset X$ be compact and let $f: K \to \mathbb{R}$ be continuous. For every open set $U \supset K$ there exists a continuous function $g: X \to \mathbb{R}$ with $\operatorname{supp}{g} \subset U$ such that $g|_{K} = f$. (tbc)

@Asaf: in which meaning do you use 'your mother'?

(continued): to see this, let $h: X \to \mathbb{R}$ be a continuous function with compact support extending $f$ and $\|f\|_{\infty} = \|h\|_\infty$. By Urysohn's lemma there exists a function $k: X \to \mathbb{R}$ with $0 \leq k \leq 1$ and $k|_K = 1$ and $k|_U = 0$ (by the version in your question). Now let $g = hk$. You will have $g|_{K} = h|_{K} k|_{K} = f$ and $\|g\| \leq \|h\| \cdot \|k\| = \|f\|$ and $\operatorname{supp}{g} = \operatorname{supp}{h} \cap \operatorname{supp}{k} \subset U$.

@Ilya What do you mean?

9:11 PM

@AsafKaragila that was exactly my question to you

@Ilya Yes.

(that should have been $k|_{X \smallsetminus U} = 0$.)

@AsafKaragila There was some discussion about what you mean when you say "your mother" to people. Does it simply mean the vulgar deprecation, or does it have some other meaning?

@AsafKaragila la repetation est la mere des etudes: in which meaning do you use 'your mother'?

@robjohn thank you ) if we are two halves of the one, then you're the wise half

@robjohn Very good. At the moment, I have no further complaints. I'll look at it a bit later.

Thanks for this!

9:14 PM

It's just a generic reply, with just a hint of vulgarity... no actual content or attempt to hurt people.

@AsafKaragila thanks - you see, if you put a bit of an effort you can create an answer

Now you'll have to excuse me. It's refill night, and I have to go get hammered.

@tb Thanks for all the criticism. I will ping Didier and see if he is happier with the modified answer.

@Ilya An answer to what?

@robjohn what are you talking about?

@AsafKaragila to either of my questions

9:17 PM

@Ilya tb has been going through one of my answers and trying to see where Didier might have thought it lacking.

@Ilya 42.

@AsafKaragila you've mixed up me with some mice

@Ilya or at least for someone who cares about cheese.

Do we have something like Springerlink for Cambridge University Press books?

@JonasTeuwen dunno ( never used it

9:19 PM

Ignore my previous comment, tb.

@JonasTeuwen btw, where would you advise to have a dinner in Brussels and Antwerpen?

Hmm. What budget and what kind of food are you looking for?

@Ilya These sort of mice?

@AsafKaragila does it eat cheese?

@JonasTeuwen just normal tasty food. the budget is average 20 pp without drinks

@JonasTeuwen the last time I was to seafood district close to Centraal Station, but non-sea food there was not too tasty. Or maybe I didn't find a good place there

20pp? :-). No I don't know anything for that.

I'll ask my brother, he lives there.

9:26 PM

@JonasTeuwen ha, never mind then. if only you knew how hard is to convince them even for that price. it's a big money in Russia

No, my brother will know.

I have mailed him, let's see if he will respond.

@tb With this version I don't need to fix anything. But in the post I'll stick to the thing imposed by the assignment.

@AsafKaragila yes.

It surely is possible, but I prefer to go less often and pay more.

@JonasTeuwen then tell me which place and how much - I will go there alone another time

9:28 PM

@Matt The problem is what I mentioned here

(so you need to do something: your version could create a huge support for the Tietze extension, so you need to cut it down in order to control the integral)

@tb I'm about to do something about it : )

9:43 PM

@tb To apply Tietze $O^c$ would have to be compact, right?

@Ilya tzilte.be :-).

It has two Michelin stars.

Apply it to $s|_{C_{\delta^\prime} \cup O^c}$ I meant to write but it was too late to edit.

@Matt yes, that's right (and that's not a good hypothesis to have!). That's why you need a bit more massaging... Actually the proof of the version of Tietze you stated should be done in such a way that you get the improved version I gave directly, but from the given tools I don't see a cheaper trick.

Can this problem solve by changing variable formula, i heard the changing of variable formula can work only up to 3 variables- math.stackexchange.com/questions/95120/…

@tb When did I state a proof of Tietze?

9:54 PM

@Matt You didn't. The proof of (the version of Tietze you stated) :). Actually I know only one proof of Tietze and the proof of the result actually shows more than what's stated in your question.

@tb Parse error : )

@tb So the fix consists of stating the non-evil version of Tietze you stated above and use that? Is this what you're saying?

Probably not. I should probably go to bed and stop badgering you. And re-read our discussion. The answer is probably there.

@Matt yes, exactly. I was looking for an online reference for that, but people don't like to state it in this form.

10:14 PM

@tb So the evil version was put on the assignment to make it more painful to solve?

: )

@Matt I don't think it was out of purpose. It's the version you find everywhere when you Google it... the point is a slightly subtle one which you only see when you go through the argument in full detail :)

@tb And that's not included in the "service" because it doesn't matter if students are miserable because (almost) no one there cares. (This rant stops here.)

@tb Thanks for today!

10:30 PM

anyone could answer my problem?

@robjohn?

@Victor yes?

@robjohn - My problem is chat.stackexchange.com/transcript/message/2879852#2879852

@Matt Well, even Terry Tao sweeps this point under the rug in his ach so wunderbaren notes that pollute all Google searches. Have a nice evening and don't be too hard on yourself: It's a minor point, really.

@tb Do I hear Tao's name?

What's swept under the rug?

@Srivatsan There could be worse ghosts haunting you.

10:36 PM

@tb What do you mean?

@Victor Do you have any indication that a change of variables will help in this problem?

@Srivatsan Oh, don't worry. I'm just having a hard time keeping a rant to myself.

@robjohn - i just want the integral actually.

Holy monkies.

@tb By the way, Tao sweeps so many "small" points under the rug that it starts to get beautiful.

10:39 PM

@JonasTeuwen Holy monkies what?

@robjohn - i heard that changing of variable formula have no geometric proof in 4D, is that true?

@tb Maybe that's what his whisky smells like: Holy monkies.

4

@tb I'm still curious. Which notes are you talking about? I'm wondering if I have read those notes.

@Victor what do you mean by geometric proof? If you mean one that you can look at, then sure, you can't really visualize things in 4D. However, change of variables works in all dimensions.

@Srivatsan terrytao.wordpress.com/2009/03/02/…

10:43 PM

@tb Thanks. Not read his analysis notes that much.

I can't believe he gives an application from comm. algebra in a real analysis course.

user image

@Srivatsan what do you mean?

I'm going to bed. Good night folks! And don't stay up too late. Thanks for today, tb.

@robjohn- in page 9 there is a elementary method, but could not extend to 4D or more, why?

@Matt Good night, Matt!

10:47 PM

@tb Sorry, commutative C*-algebra. He proves the commutative Gelfand-Naimark theorem as an application of "the theory we have developed in the previous lectures"

@robjohn -Sorry, the image's quality is low, may be you could search From Euler to cartan on the internet.

@Srivatsan I see. Yes, impressive.

@Victor I don't see a method on page 9.

@robjohn - google.com/… sorry, it is on p.10

(serious question) Aren't there good Maple forums online?

10:55 PM

@robjohn - is there a elementary way to prove the result by induction once i have it by geometric way in 2d and 3d?

@Srivatsan All I'm saying is RTFM!

@tb I guess math questions related to Maple is on-topic, but I agree Maple questions related to Maple should be off-topic. Can you post a comment though, tb?

@Victor The lack of geometric method seems to me to be a lack of results at that time. The same equation using determinants holds true in higher dimensions.

Thanks, tb.

@robjohn - if higher dimensions is not call geometric, then what does it call for the subject related to it?

11:07 PM

@Victor It's not like higher dimension means not geometric. The point is that at that time, people knew how to proof the formula for 2 or 3 dimensions whereas 4 was difficult (I don't know why).

@Srivatsan They didn't even think about higher dimensions. Laurent Schwartz said that he didn't learn about more than three dimensions until well after he graduated.

@tb "The code I type is math related"

:)

@tb That's good to hear.

Gtg, bye!

@Srivatsan Yeah, I'll soon ask a cooking question because I think about math while eating...

@robjohn - What geometric method would recommend to use for higher dimension

@Victor Do you know the formula for finding the volume of a parallelepiped?

11:12 PM

@robjohn - i mean higher than 3D

@Victor and its extension to higher dimensions?

@robjohn - Yes

@Victor Then what is the question?

@robjohn - How to prove the higher dimention(>3) of changing of variable formula with geometric way?

At "infinitesimal" scales, all the change of variables are linear maps, and the relationship of the differentials of volume are by the determinant of the Jacobian

@Victor What do you mean by "geometric way"?

11:15 PM

@robjohn - The intuition by Geometric figures.

@Victor What does a geometric figure look like in $\mathbb{R}^4$?

@Victor We have to abstract things to move to higher dimensions.

@robjohn - i mean i can't see the intuition even with "infinitesimal" scales...

@Victor You cannot hold a 4-dimensional object in your hand or perhaps even in your mind which is designed to deal with 3-dimensions.

@Victor Do you see that the volume of a parallelepiped in $\mathbb{R}^n$ is the determinant of the vectors of the edges incident on a given vertex?

@robjohn - i mean if i want to parametrize the curves(>3) by different variable, then i get higher dimention changing of variable formula

@tb My bottle was empty.

11:20 PM

@JonasTeuwen I'm sorry; I probably finished it off :-)

@robjohn what do you mean?

:-)

@robjohn, what do you mean by the volume of a parallelepiped in R n is the determinant of the vectors of the edges incident on a given vertex?

@robjohn - Is there a article for what you are saying?

@robjohn - i see the formula in 3D, but couldn't go further...

are you familiar with the matrix form of the chain rule in $\mathbb{R}^4$?

$
\begin{bmatrix}
\frac{\partial u_1}{\partial x_1}&\frac{\partial u_1}{\partial x_2}&\frac{\partial u_1}{\partial x_3}&\frac{\partial u_1}{\partial x_4}\\
\frac{\partial u_2}{\partial x_1}&\frac{\partial u_2}{\partial x_2}&\frac{\partial u_2}{\partial x_3}&\frac{\partial u_2}{\partial x_4}\\
\frac{\partial u_3}{\partial x_1}&\frac{\partial u_3}{\partial x_2}&\frac{\partial u_3}{\partial x_3}&\frac{\partial u_3}{\partial x_4}\\
\frac{\partial u_4}{\partial x_1}&\frac{\partial u_4}{\partial x_2}&\frac{\partial u_4}{\partial x_3}&\frac{\partial u_4}{\partial x_4}

@robjohn - it is missing a $, so i couldn't see it...

11:30 PM

@Victor is there a "view full text"?

@robjohn - The mathjax isn't loading on that page even i clicked the link"render Mathjax" which is on my favorite

Did you click the "see full text" link?

@robjohn - On a seperate tab, it doen't work...

user image

If you click "(see full text)", and then the MathJax link, you should see that

@robjohn - Anyway, i am not familar with the chain rule in matrix form and matrix computation

11:39 PM

@Victor there is your problem then. You really need to be familiar with matrix operations and the chain rule and determinants to understand the abstraction from $\mathbb{R}^3$ to $\mathbb{R}^n$

@Victor There are preliminaries.

@robjohn- i understand the chain rule, but i just can't see how it related to my problem.

@Victor The chain rule is the basis of the change of variables formula.

@robjohn - how do you know that dudv=-dvdu in the multiplication?

@Victor this is usually detailed in differential forms. It is due to the fact that when you swap columns (or rows) in a matrix, the determinant is negated.

@robjohn - is there intuition on that?

11:50 PM

@Victor you're computing a signed volume depending on the orientation. Swapping two rows or two columns changes the orientation

Think of the standard coordinate system in $\mathbb{R}^3$. If you exchange the $x$ and $y$ coordinates, you change the handedness of your system (that is, its orientation).

@tb - why does the orientation change as you exchange the x and y?

@Victor because $e_x \times e_y = e_z$ while $e_y \times e_x = - e_z$.

user image

@Victor: there are two ways to change orientation.

@tb - now i get it, but how does that relate with 4D?

@Victor: swap variables or change the direction of one.

11:58 PM

@robjohn - Thanks but i think i have no hope to solving higher dimension without the "figures".

@Victor $\det \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} = 1$ while $\det \begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} = -1$

@Victor Those figures I posted are 2-dimensional slices of the 4-d situation.

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