Mathematics - 2012-08-29 (page 1 of 2)

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12:00 AM

@FortuonPaendrag 2 days left. not much time...

Fortuon Paendrag

uh oh.

Moved the Q: math.stackexchange.com/questions/188151/…

Gustavo Bandeira

12:38 AM

Guys. [Look this](http://pt.wikipedia.org/wiki/N%C3%BAmero_de_Euler#Caracteriza.C3.A7.C3.B5es_menos_triviais_de)

$$\frac{1}{0!}$$

@GustavoBandeira Scary.

Gustavo Bandeira

@JonasTeuwen This isn't viable, right? Unless mathematics is different in Brazil....

Is it possible to add some self-programmed annoucements to chat -room?

Gustavo Bandeira

0! = 1?!?!

WTF?!

Moved it to META here.

1:03 AM

@GustavoBandeira Surprise!

@GustavoBandeira Also, your pic qualifies for a Quentin Tarantino movie.

Gustavo Bandeira

@PeterTamaroff Yes.

It has something to do with analytic continuation

@PeterTamaroff Yep. I love this picture. =D

@GustavoBandeira You can probably give it a combinatorial interpretation.

Gustavo Bandeira

@PeterTamaroff It transmit the YGGR feeling

@GustavoBandeira HAHAHHAHAHA

Gustavo Bandeira

@PeterTamaroff Tell me more.

1:07 AM

@GustavoBandeira Looks kind of psycho, yes.

@GustavoBandeira Well, given a set $S$ with $n$ disctinct elements, the number of permutations of the elements of $S$ is $n!$, right or left?

Gustavo Bandeira

Mine picture still have more style, it's kinda YGGR and I'm laughing!

@GustavoBandeira *My

@GustavoBandeira "My picture has still more style"

Gustavo Bandeira

Yep. To use "mine" I guess I would have do say "Mine still..."

=P

Gustavo Bandeira

@PeterTamaroff Right.

1:11 AM

@GustavoBandeira The idea now is to interpret $0!$ as the number of permutations of elements of $\varnothing$...

Gustavo Bandeira

But the empty set has no elements, isn't it?°

@GustavoBandeira Yes. It has no elements. You can define it, for example, with $\varnothing=\{x:x\neq x\}$

Gustavo Bandeira

So... how is it possible to have permutations with the elements of the empty set?

@GustavoBandeira Well, I'm thinking it as the "state" of the set. For example, $2=\{0,1\}$ has $2!$ "states"; $2=\{1,0\}$, and $2=\{0,1\}$ Now, say I have a pefectly homogeneous circle. If I draw 2 symmetryc points on it I get $2!=2$ symmetries. If I have one dot on it, I get only $1!=1$. If I draw 3 points at 120 dgs on it I get (I'm imagining) I can get $3!=6$ configuations (think about "flipping it over, too"). Now if I have no points, I get only one "state", right?

Gustavo Bandeira

@PeterTamaroff Thinking.

@PeterTamaroff Yes

1:23 AM

@GustavoBandeira Well, that's my idea.

Gustavo Bandeira

I guess I got it, then $0=\{\varnothing\}$ which is still have 1 "state"

@GustavoBandeira No no no. $0=\varnothing$

Gustavo Bandeira

I guess I got it, then $0=\{\varnothing\}$ which still have 1 "state"

$1=\{0\}$

Gustavo Bandeira

But both 1 and 0 share the same number of states, right?

1:26 AM

@GustavoBandeira Yes, but that doesn't mean they are the same set.

Gustavo Bandeira

Yep. 1 = 1 set with a empty set. 0 = The empty set.

2:06 AM

@gustavo: consider the possible lists you can make with the elements of the set $\{1, 2, 3\}$. There are 6 possible lists: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1), right?

Gustavo Bandeira

@MJD Yes.

6, because 6=3!, right?

Gustavo Bandeira

Yes.

How many lists can you make with the elements of the set $\{1,2\}$?

found a typo in the title of a paper

Gustavo Bandeira

2:17 AM

2!?

Can you tell me what they are?

@anon that's gotta be embarrasing!

@anon what is the typo?

I've made typos in email subject lines, though, so I sympathize with the typist.

bateman-home

?

2:27 AM

It's Horn (as used correctly in the paper), not Horm (as used in the title)

uhmmm, horn like:

user image

2:52 AM

Anyone interested in closing an exact exact duplicate math.stackexchange.com/questions/188187/… ?

Gustavo Bandeira

@MJD Sorry, I had to solve a problem.

OK

Gustavo Bandeira

@MJD The are combinations?

I don't know what you are asking.

If you can make 6 lists from the elements of $\{1,2,3\}$ (and I showed the 6 lists), how many lists can you make with the elements of $\{1,2\}$? You said 2. What are the 2 lists then?

Gustavo Bandeira

{1,2} and {2,1}

They are combinations *

3:07 AM

Okay. How many lists can you make with the elements of $\{1\}$?

Gustavo Bandeira

1

What is the 1 list?

Gustavo Bandeira

{1}

How many lists can you make with the elements of $\{\}$?

Gustavo Bandeira

Should I consider the axiom of the empty set?

3:12 AM

You should not consider any axiom at all.

You should just answer the question.

Time is up. Good night!

user image

${}$

Hey, that's a message though it's empty!

you've discovered the chatjax bookmark I see

Gustavo Bandeira

@MJD Well, If I consider only the elements - what is inside the {} - I guess there's no element at all. If the list must exist, then there's one list.

@peoplepower Are you trolling me?

I did not intend to troll.

Mariano Suárez-Alvarez

3:18 AM

it just happened naturally?

I leave it to you to determine whether it occurred or not.

Gustavo Bandeira

@peoplepower You leave it to the reader as an exercise?

Though I don't know the answer.

1 hour later…

4:32 AM

Clarence? What are you doing here?

Gustavo Bandeira

4:43 AM

@yunone ?!

yu speaks of Clarity Clarence, the name of the meme I posted just above

Gustavo Bandeira

Oh, ok.

Are there other problem solving websites like Project Euler?

I've hear on polymath project by Gowers.

5:03 AM

Morning. x_x

hi!

5:24 AM

Heh. Did anyone see this?

5:35 AM

tsk tsk

6:09 AM

@anon Did you also read the angry comment about "I edited the question so that my answer is now correct, QED"?

@Matt I cannot believe someone has not given an answer yet. Is it because the OP has specifically asked for hints?

By answer, I mean the formal answer, not the one by Andre Caldas in the comments.

hi all

anyone here??

@experimentX aye

can you help to to evaluate this integral?

@experimentX depends on how clumsy I am today. But no harm in trying I guess.

6:24 AM

$$\int_0^\infty e^{-(x-a/x)^2}dx$$

You missed a space between \infty and e

just hint ... for any numeric value of 'a' it seems to have value sqrt(pi)/2

why?

here a link on WA

Integral does not converge it says

6:30 AM

sorry ... could you change that 2 into ^2

done, I can see it converges, I just cannot find the appropriate transformation right now. I am highly suspicious this is not a closed form.

Also you will have to use the small e

I am trying to convert it into $\int e^{-x^2} \mathrm{d}x$

I'm also trying that ...

it's very bad :((

I was thinking may be we can apply by parts?

Separate out all the three terms in $\left(x-\frac{a}{x}\right)^2$

Or even better might be to apply differnetiation under the integral sign

oh the original problem was stated as $ e^{-x^2 - a^2/x^2}$

ohh

@WillHunting Can you please see the above problem we are trying to solve?

user19161

6:50 AM

@JayeshBadwaik Nope, I am leaving now, bye.

@WillHunting bye

@experimentX Do you know the polar method of evaluating $\int e^{-x^2} \mathrm{d}x$

2

A: Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge?

It does, and you can also compute its value: $\int_{[0, \infty) \times[0, \infty)} e^{-(x^2+y^2)} dx dy = \int_{[0, \infty)}\big(\int_{[0, \infty)} e^{-(x^2+y^2)} dy \big)dx = \int_{[0, \infty)} e ^ {-x^2}\big(\int_{[0, \infty)} e^{-y^2} dy \big)dx =$ $= \int_{[0, \infty)} e ^ {-x^2}dx \ \ \int_...

yeah i know ...

i tried to mess around with that ... and i get infinite radius of convergence

Ohh. Hmm. I will work on this then, will let you know if I get anything.

However I am thinking, it might be easier to solve the problem in the completed-square form that you have originally stated

7:05 AM

if i get ... with this infinite series ... then i'll get the other one

Hi all

Hi

Could any help me understand what is wrong with this solution? (math.stackexchange.com/questions/171970/…)

@experimentX I was thinking may be we can do like this
$e^{-(x-a/x)^2} > e^{-x^2}$ for all $x$, correct?

i think yes ...

looks like people don't like reading long stuff :)

7:14 AM

Then, let $f(x) = e^{-(x-a/x)^2}$ and $g(x) = e^{-x^2}$

Now try to calculuate the limit
\begin{equation}
\lim\limits_{x \rightarrow \infty} \frac{f(x)}{g(x)}
\end{equation}

If we can show this is constant, then either both integrals converge or both diverge.

tends to 1

both function tend to each other at infinity ... but w are integrating

@experimentX limit comparison test of improper integrals

shows that it converges right??

Yup.

And from the part that $f(x) > g(x)$, we know the integral must converge to a value greater than that for $g(x)$

Actually, we should have assumed $g(x) = e^{2a} e^{-x^2}$

The above results will still hold and we get a lower bound on the integral

and now if we can find an upperbound and then try to sandwich, it might be possible.

@experimentX can i get the source of this question?

openstudy.com/study#/updates/503d76a2e4b04d6bed310c8f

7:21 AM

@experimentX: do you think my proof is too long? :)

haha ... I would't read that long. probably many won't

@Chris'sister considering that others have written a very shorter version yes. Also, while you have used all the gamma stuff, one solution has a very simple substitution.
But that is not a reason for downvote I would say, it is always important to know as many methods as possible,

The downvotes might have been because of the last comment by JM and it is possible that the downvoters will remove once they see that you have corrected it.

@JayeshBadwaik: notice that my way uses a real method.

I doubt one can obtain a solution much more shorter.

@Chris'sister as I said, not a real reason to downvote and one of the downvotes has now been removed

@experimentX I guess we got the inequalities the wrong way around

$f(x) < g(x)$

with my new $g(x)$ that is

so we have a upper bound

now we want a lower bound

so you are planning to sqeeze it in that value?

7:29 AM

yes

I wish there is a more elegant method

but then what you have is what you have

also, we should post it on main I think

seems non-trivial to me

@JayeshBadwaik all right I'm posting

damn quality standards

quality standards? of the site? or is it from where the integral came from?

of the site ... I don't like typing too much

here the link math.stackexchange.com/questions/188257/…

you might want to add the tag improper-integrals

there, you have it now. Not so difficult now that we see it.

7:49 AM

I'm such an idoit

@experimentX yeah, tell me about it, nothing different here

Hi folks

@OldJohn Hi, just going to lunch, talk to you later

hi @OldJohn

@JayeshBadwaik Later :)

@experimentX Hi

7:53 AM

@experimentX from where was this problem?

@OldJohn by your experience, how obvious is this solution?
feeling gutted about not begin able to see it
http://math.stackexchange.com/a/188260/14082

he just changed $x - a/x = z$

@JayeshBadwaik I would have said that the first step is not obvious, but the substitution is a pretty common idea - I would have looked at that first

$ dz = 1$ + a/x^2$ dx $

clever manipulations

I saw that f(x)=f(a/x) and tried to split the integral.

I did try the substitution, did not succeed, may be I made some error on the way and gave up.

I need more practice.

8:00 AM

I need more practice too

I just flagged this for deletion as "not welcome in our community".

@Matt agreed - does it need more flags?

@Matt Agreed.

@OldJohn Possibly. I'm not sure.

I'm angry how garbage like this could stay in the thread without being purged for so long.

@Matt yes, the OP is seriously in trouble and such kind of things can put any ideas in his mind.

It is on the brink of being dangerous.

8:07 AM

see you guys ...

@experimentX bye

@JayeshBadwaik thank you very much for your time

i'll be back
If i have more problems

@experimentX sure, I would like to confront as many as possible .

I have just been entertained by this question the poster says of one answer that it is ultimate and excellent ... and then accepts a different answer!! (and doesn't seem to fully understand either answer)

2

Perhaps the accept is a poor way to draw attention to her comment that she needs more help?

8:12 AM

Where is that link about making no hurry to accept answers?

For a given structure of a quantified theory, is the valuation (en.wikipedia.org/wiki/…) of any formula (or at least sentence) always practically executable? It's known that there isn't always a proof, but does this relate to the valuation? In the case of propositional calculus, where every formula has a certain degree, this is the case as you just climb down to atomic propositions. I don't know if this is also as easy for predicate logic.

If it's true that for a given structure you can automatically evaluate it, then I don't see why finding a proof (by which I mean writing down the sequence constructed in say a Hilbert style system) or rather the possibility not to be able to find it would be such a big of a deal. To do the valuation seems like a proof in a way. On the other hand, I don't know where the problem with the evaluation could kick in (maybe related to there being more possible variables for a formula).

Valuation (logic)

In logic and model theory, a valuation can be: *In propositional logic, an assignment of truth values to propositional variables, with a corresponding assignment of truth values to all propositional formulas with those variables. *In first-order logic and higher-order logics, a structure, (the interpretation) and the corresponding assignment of a truth value to each sentence in the language for that structure (the valuation proper). The interpretation must be a homomorphism, while valuation is simply a function. Mathematical logic In mathematical logic (especially model theory), a valuati...

1 hour later…

9:35 AM

How can I see the questions I have starred earlier?

@JayeshBadwaik In your list of favourites?

(or favorites in American)

@OldJohn Okay. Do I have to go to my profile for that? Or is there a shorter way?

Anyway, I have bookmarked the link now.

@JayeshBadwaik I only know how to get there via the profile page - but if there is something I want to visit more quickly, I bookmark it in Chrome

@OldJohn Thanks . I am getting dumber day by day. I guess I should take a break to freshen myself up.

@OldJohn Don't be angry with me for my gaffes.

Good morning everybody!

9:44 AM

@JayeshBadwaik I need a break too ... and yes, I am an old man ... so no need to apologise :)

and I almost never get angry - don't see the point

@Nimza Good morning

Is it possible to find 3 pluriharmonic functions in $\mathbb{C}^2$ such that $\frac{\partial f_1}{\partial f_0}$ and $\frac{\partial f_2}{\partial f_0}$ are holomorphic? $\partial f = \sum_{j} \frac{\partial f}{\partial z_j} dz_j$

@OldJohn hi!)

I consider forms on Riemann surface

Darn - Linux locked up and needed reboot. Its not supposed to do that :(

10:00 AM

What distro?

@peoplepower Latest version of Mint - the only one I had lying around when I installed recently

10:28 AM

@OldJohn Do you think there should be some more information about the metric spaces in math.stackexchange.com/questions/188289/…?

I think one of the metric spaces should be complete or something.

@JayeshBadwaik I was wondering about that question too - but I am not sure at the moment, too busy helping my wife to give it full attention :(

@OldJohn No problem. I will just type out what I think. You can see it later. I think the fact that $f(X_{i})$ is compact implies there is a finite open cover and hence, this finite open cover is pulled back to a finite open cover in $X$ due to continuity and then the union of the finite open cover in X is again open because of finiteness. Something on this lines has to be there. I cannot make it precise though.

@JayeshBadwaik OK - will take a look later

user19161

10:51 AM

@OldJohn Mint is getting very buggy these days, like Ubuntu.

11:20 AM

@WillHunting Yep - and I am also unsure about my hardware, as I seem to have had an increase in problems since I installed an SSD as my OS drive

@OldJohn what kind of problems?

@JayeshBadwaik Lock-ups and crashes (e.g. Chrome "Aw Snap") - not often, but a bit annoying

11:37 AM

I now asked my question above in on the main site

@OldJohn youtube.com/watch?v=hUKBuAkr4Lg

@BenjaLim interesting!

yes

12:06 PM

I here, Helsinki there 8-(.

@JonasTeuwen Why do you want to go to Finland?

If I remember correctly, Eero Saksamann will be in Helsinki.
https://wiki.helsinki.fi/display/mathstatHenkilokunta/Saksman,+Eero
I sent him an email once.

I was going to be at a conference there.

@JonasTeuwen then?

I AM ILL.

Okay, back to bed. Bye.

12:12 PM

later

Later

latest is I just got 2 rep points deducted from account without any downvote or anything. must be some deletion.

Must be.

Henning Makholm

@JayeshBadwaik Do you have the "show removed posts" on your rep page checked?

@HenningMakholm Oh. I did not see the checkbox before, just enabled it. sure enough there is an answer deleted which I edited a few days back.

1:09 PM

I need some linear algebra help.

Does anyone in here fancy doing some linear algebra?

I have a vector space $V$ and a bilinear form $b: V \times V \to Z_2$.

I have picked a basis for $V$, $e_1, \dots e_n, f_1, \dots , f_n$ such that $b(e_i, e_j) = 0 = b(f_i, f_j) $ and $b(e_i, f_j) = \delta_{ij}$.

Now I want to compute the determinant of the matrix representing $b$ in this basis.

Duh.
The matrix looks like $I_{2n}$, right?

@Matt Why do you have two sets of basis?

Otherwise, how do I specify the properties I specified for $b(,)$?

No it does not look like $I_{2n}$.

but would its dimension not be $n \times n$ ?

$V$ has dimension $2n$ so I need my matrix to have dimension $2n \times 2n$.

Ohh. I thought V has dimension $n$.

Sorry, now got your problem correctly.

1:20 PM

The form looks like $b(x,y) = x^TAy$.

So $A$ will contain more zeros than $I_{2n}$.

No. Grrrrrrr.

Why do I find this confusing?

It is a banded matrix. Divide the matrix into four parts.

The top right, and the bottom left parts will be indentity matrices.

and the other two will be zero matrices.

I hope you get what I am saying.

Yes that's what I thought at first but it must be wrong since the determinant should turn out to be $1$.

And yes, I understand what you're saying. : )

I am very bad at explanation hence asked. :-)
The determinant is 1.

Duh.

@JayeshBadwaik Yes, thanks. It is one.

I should learn linear algebra.

Anyway, thanks a lot and see you later!

Later.

1:26 PM

@Matt I know

how to compute the simplicial homology of the klein bottle!

If $f$ is a holomorphic bijection, with nonzero derivative, why is $f^{-1}$ is holomorphic? Is it just the inverse function theorem?

1:41 PM

Yes.

thanks

Henning Makholm

2:02 PM

@QualFighter I don't think you can even have a holomorphic bijection that has zero derivative anywhere. (It wouldn't be bijective).

(Unless defined on a non-open set, and the zero derivative appears at a boundary point -- to the extend this is allowed by the details of your definition of "holomorphic").

@anon ah! - the old, recognisable Anon is back :)

I was thinking of a question (about a possible connection between an identity of Ramanujan and zeta functions of groups) and considered putting it on MO, in which case I would do better with a more professional gravatar. I decided I'm going to put it on MSE first and see what happens.

BTW $p$-adic methods appear to be very useful in ZFoGs, and these zetas are generalizations of your usual number-theory zetas. I thought that was interesting.

I'm in my algebra class here learning about the cross product and stuff

Isn't cross product from vector analysis? Or do you mean tensor product and abstract algebra?

@anon definitely sounds interesting - although I don't think I have made enough progress with $p$-adic analysis (yet) to understand the likely implications (but I am working on it ... slowly)

2:15 PM

I understand it's hot out, but this building takes air conditioning way too far.

I will soon become iceman anon.

It is algebra. So it has some vector algebra in rit

My teacher says he's interested in forming a p-adic analysis class next semester. I hope he follows through.

thunder, lightning and lots of rain here - no need for air con.

@anon That would be great - who is the teacher? anyone famous?

this guy. I don't know that he's particularly famous.

@anon Probably not - but that is not terribly important. $p$-adic analysis is fascinating, though - from the little I understand so far

2:22 PM

@PeterTamaroff You may be interested to learn that the only dimensions of vector spaces a binary cross product may live in is in dimensions 3 and 7 (this fact is related to the quaternions and octonions and classification of division algebras, I think). There are also unary cross operations in even dimensions, a terny cross operation in 8 dimensions, and a (geometrically obvious) $d-1$-ary cross operation in $d$ dimensions. See: unizar.es/matematicas/algebra/elduque/Talks/crossproducts.pdf

Instead of buying the book, I have thus far only copied the homework questions from the book literally while skimming it in the bookstore (for my algebra class). I wonder if that counts as real-world piracy.

How much does the book cost?

I think the lowest used price is around 60-80 online (after trickily horrendous shipping prices), which is pretty modest.

I haven't got around to it yet.

So, about 200 brand new?

yes

@HenningMakholm I hope your dog is now suitably smooth - and presumably ironed in conformity with the hairy ball theorem

2:34 PM

Hairy ball theorem.

@skullpatrol :)

@OldJohn ;)

Casorati-Weierstrass shows that near essential singularities of $f$, the image of $f$ is dense in $\mathbb{C}$. What can we say about the image of $f$ near poles? Obviously the image is unbounded, but can it be dense? When is it not?

Are my questions even good questions?

In mathematics the art of asking questions is

more valuable than solving problems.

Cantor (1867)

So are my questions lacking artistry?

2:46 PM

Like any art, it requires practice.

are you making fun of me, skullpatrol?

no

ok

Henning Makholm

@QualFighter By definition, near a pole the function behaves more or less like $z^{-n}$. So the image of a small enough punctured neighborhood of the pole will be the complement of a (larger and larger) neighborhood of the origin.

@HenningMakholm, thanks, that makes clear sense

3:02 PM

Why don't you look it up in a book or on the internet?

Henning Makholm

@Matt This is the internet.

@anon that's pretty dope to know

Henning Makholm

@OldJohn Are you assuming a spherical dog here?

@HenningMakholm Yes. But it's not looking it up.

And also well done on that looting!

3:13 PM

@PeterTamaroff What "looting"?

Oh, the "piracy."

@anon awsuuuum paper, though I can't really take much out of it yet!!!

@skull yeah that =)

Henning Makholm

3:36 PM

Call for meta participation

Michael Greinecker

3:50 PM

@HenningMakholm I just signed up for that intimidating place.

What's this? I've never seen it before.

Michael Greinecker

Meta Stack Exchange

Henning Makholm

@ZhenLin What's what? The review interface was revamped some weeks ago if that's what you're asking.

Ah, a new feature then...

Michael Greinecker

Apparently, I have insufficient rep to downvote there.

4:11 PM

lol Riemannian zeta function

I want to show that if $f^2$ and $f^3$ are analytic, then so is $f$. I want to say that $f^2+f^3=f^2(1+f)$ is analytic, and thus $1+f$ is, and so thus $f$ is, but I think this is weak on the details. Am I on the right track?

the quotient of two analytic functions is analytic wherever the denominator is nonzero right?

Gustavo Bandeira

I've made a question can you give me opinion on it?

I see, you are suggesting I see $f$ as $f^3/f^2$, and show that it cannot have any poles

well, the second can have removable singularities (wherever $f$ is zero), but after removal we are left with just $f$

(and the process of removing isolated singularities to extend an open domain should preserve analyticity)

there's a special case of $f$ identically zero you technically need to get out of the way first

4:18 PM

thanks @anon. so is my argument wrong?

if $f\cdot g$ is analytic, and $f$ is analytic, can $g$ not be analytic?

i guess your argument works here too

i was thinking in terms of taylor series and such

no, it can't be, but in order to show that you'd have to use the same argument, so there are superfluous steps

(meaning, it can't not be analytic)

Hello,everyone!

Gustavo Bandeira

@MeAndMath Yo!

Good?

how r u?

Gustavo Bandeira

Fine, ya?

4:33 PM

fine.

Gustavo Bandeira

I'll go out to lucnh.

If 3G is fine, we will keep chatting.

Cya (or not)

As you wish.

4:45 PM

Hi @MarianoSuárez-Alvarez!

Mariano Suárez-Alvarez

Hello :-)

Michael Greinecker

Hello @MarianoSuárez-Alvarez

¿bien?

Mariano Suárez-Alvarez

todo bien

que bueno

what about your office?is it done?

Mariano Suárez-Alvarez

4:56 PM

yeah :-)

half-done, really, because it'll be painted sometime soon

so I have most things in boxes

but oh well :-)

Where has WillHunting disappeared?

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