Mathematics - 2012-05-06 (page 3 of 5)

@DavidWallace You'd be surprised. There were about 100 kiwis where I live that were stranded and could not go home when some volcano exploded last year

David Wallace

The Chile one?

Benjamin Lim

11:06 AM

yeah in june last year

David Wallace

But they weren't stranded for long, right? I never thought of Canberra as being a hot item for kiwis to head to. BTW, I looked at your profile. It says you're in Seednee.

Benjamin Lim

@DavidWallace No that is not technically correct :D :D :D

Look at my supervisor's CV

David Wallace

OK, so you're at ANU, but your profile still says Seednee. Are you deliberately trying to confuse the masses?

Benjamin Lim

@DavidWallace What did my supervisor's CV say?

David Wallace

ANU.

Benjamin Lim

11:13 AM

@DavidWallace Well he's at ANU I could be somewhere else right?

David Wallace

Well, anywhere really. What's the point of this?

Benjamin Lim

@DavidWallace Sorry, yes I am at ANU.

I did not mean to beat around the bush or anything

David Wallace

I know that there are huge numbers of kiwis in Brizza, Melbourne and Seednee. I didn't think we bothered with the smaller cities too much.

Benjamin Lim

@DavidWallace It's because of the uni

David Wallace

Do we have to pay full overseas-student tuition fees? Or is it cheaper for kiwis?

Benjamin Lim

11:32 AM

@DavidWallace You guys get Commonwealth Supported Places, but I believe have to pay upfront

David Wallace

OK, I'm just trying to work out why so many of us would come to you.

Benjamin Lim

dunno

Matt N.

NARQ.

Asaf would like Brian/Teddy/JM/and others to look at this.

skullpatrol

11:58 AM

You're right he will never get bored of this ;-)

Zhen Lin

I find the discussion quite amusing. It seems to boil down to a fundamental disagreement about definitions.

Nimza

12:23 PM

Hi. Does anybody know a global inverse function theorem in positive ortant?

J. M.

12:45 PM

That question on counterintuitive results is now at thirty answers. I'm sorely tempted to post the thirty-first for giggles.

robjohn

@JM I see 29, but go for it :-)

J. M.

@robjohn Ack, I forgot that the numbering counts deleted answers. I'll wait for the thirtieth answer first... >:)

robjohn

@JM It does for those who can see deleted answers :-)

J. M.

I don't quite like the way this question is composed.

Benjamin Lim

@JM voting to close

robjohn

12:59 PM

@BenjaminLim Might it be seen as "History and development of mathematics"?

@Américo: Good morning (?)

Good intended - questioning the morning-ness :-)

Américo Tavares

@J. M. How did you come with the idea of computing the even part of the Tito's c.f?

@robjohn Here is good afternoon!

@J. M. was it just to speed up the convergence rate?

@JM Has Mathematica some built in computing facilities to allow obtain the results of transformations to a c.f?

J. M.

Hi @Américo! Yes, I was looking to accelerate the convergence of the CF for computational purposes.

robjohn

@Américo: I have never seen the notation $\underset{n=1}{\overset{\infty}{\mathbb{K}}}$ before. Is it common?

J. M.

No, Mathematica does not have the facility for CF manipulation. I did the whole thing by hand.

@robjohn In CF literature, yes. It's attributed to Gauss.

(...and transcribing what I wrote on paper to $\LaTeX$ was... rather exhausting)

Américo Tavares

That notation is called Gauss Notation for generalized continued fractions

J. M.

1:11 PM

Again, thanks for the nudge! I had completely forgotten what the CFs for odd zeta values looked like.

I'll probably edit the answer later to include the derivation of the odd part for completeness.

Américo Tavares

@JM You are welcome! I thought of a convenient Bauer-Muir transformation might do as well, but I didn't try enough.

t.b.

@MattN it's gone.

Américo Tavares

@JM If the proposal for a blog gets accepted this question and your answer could give a good post, I think.

J. M.

Well, I'm happy with writing on my current blog, but I suppose... :)

Américo Tavares

@JM What is your blog?

Nimza

1:25 PM

If an open subset $M$ of $\mathbb{R}^n$ is simply connected and $f \colon M \to \mathbb{R}^n$ is continuous but not injective will $f(M)$ be simple connected?

t.b.

@Nimza think about the exponential map :)

kahen

I'd suggest a projection down to $\mathbb R^2$

Nimza

@tb exponential map -- it's from differential geometry? I didn't deal with it :(

t.b.

@Nimza what's the image of $z \mapsto e^z$ from $\mathbb{C} \to \mathbb{C}$?

Nimza

@tb ah, thanks!

t.b.

1:32 PM

@Nimza the point of my example is: as soon as you have a covering map the image will have a nontrivial fundamental group.

robjohn

@AméricoTavares Marvelous! I have written a paper (sort of an ongoing project) starting in high school on CF in which I attribute that recurrence (in the more restricted setting) to Wallis. Now I have a reference. I also named this algorithm partly for him because of that recurrence.

J. M.

@AméricoTavares tpfto.wordpress.com is the address. I believe you've been to it before...

robjohn

@JM JM is watching :-)

J. M.

The three-term recurrence is nice for symbolic manipulation. For numerics, however, I like Lentz-Thompson-Barnett.

Nimza

@tb I don't know such terms :( But I understand example, thank you

Américo Tavares

1:36 PM

@robjohn I will read your paper.

Nimza

@tb I have only to apply the global inverse function theorem and then I have to test if the image in simply connected :)

Américo Tavares

@JM Yes, I now think so. It was when your name was written in full here, I think.

@JM I even follow your blog!

t.b.

@Nimza Yes, the global inverse theorem (in its easiest form) gives you that a locally injective smooth map $\mathbb{R}^n \to \mathbb{R}^n$ satisfying $\lim_{|z| \to \infty} |f(z))| = \infty$ is globally invertible, hence the image will be simply connected. Note that the exponential map is a counterexample showing this "properness condition" is necessary, as $|2\pi i n| \to \infty$ while $\exp{(2\pi i n)} = 1 \nrightarrow \infty$.

J. M.

@AméricoTavares Ah, a very long time ago...

Américo Tavares

@JM so it was.

@JM What should I do to get my question on the Apéry's divergent series answered in such a way I can understand?

J. M.

1:52 PM

@AméricoTavares Let me look...

It might take me a while to review, since I need to brush up on the Apéry route... and Wim's answer isn't really an answer.

Gigili

@robjohn What a great paper.

Américo Tavares

@JM Thanks. That's what I think. I tried to ask more details, but he didn't provided anything significant to me. I posted it on MO too, but I made a mistake. I've upvoted an answer just providing the link to the well known Alf van der Poorten's paper. As a result it is no longer listed in the answered questions.

t.b.

@robjohn: would you agree that this is a duplicate?

(I guess I prefer to wait for the OPs reaction)

J. M.

@AméricoTavares I'll have to review how Pochhammer symbols behave in series like those. "Factorial series" is the term I've heard for these, but I must confess I don't yet have a good grasp of their convergence properties.

N3buchadnezzar

@tb Esxcuse me for being silly again but what does the notation $g \in L^2[0,1]$ mean?

Does it mean a square? ..

t.b.

2:00 PM

@N3buchadnezzar $\int_{0}^1 |g|^2 \,dx \lt \infty$

N3buchadnezzar

Thanks, never seen that type of notation before with the exponent.

Américo Tavares

JM I assumed that since Apéry stated that expansion on his short paper, that should be known at that time.

t.b.

$L^2$ is the space of square integrable functions. Similarly, $f \in L^p$ means $\int |f|^p \lt \infty$ for $1 \leq p \lt \infty$

Gigili

@tb: Is it okay to edit a question to fix the grammatical errors?

t.b.

@N3buchadnezzar: see also here

@Gigili If there are enough of them, I'd be all for it :) Yes, I think so.

J. M.

2:04 PM

@AméricoTavares I suppose...

Gigili

@tb Define "enough", a million errors?

t.b.

@Gigili I'd have said: as soon as you don't need to cheat the engine to submit the edit suggestion, there's enough of them :)

Gigili

@tb Thank you, schatz! I was trying to avoid another unanimous rejection.

Matt N.

Is this right?

Never mind, I guess I'll find out soon enough...

J. M.

I have a good feeling this is a dupe...

Matt N.

2:17 PM

@JM I wouldn't know -- I never look at thingies tagged special-functions.

"measere"? Sometimes I really wonder how these people use their keyboards to type in their questions.

t.b.

@MattN how do you know from $\int fg \lt \infty$ for all $f$ that $\sup_f \int fg \lt \infty$?

N3buchadnezzar

@Gigili If the meaning of the post is unclear because of spelling fix it, otherwise leave it be =)

Matt N.

Hm...

t.b.

I believe there's no way to avoid the uniform boundedness principle or mimicking its proof, as robjohn does in the thread I linked to.

Matt N.

Had a look but didn't read.

kahen

2:21 PM

math.stackexchange.com/questions/141768/… --- this question paints a pictures of extreme mathematics education sadness. How could it be possible to know of totally disconnected spaces and not know of the Cantor set?

Matt N.

@kahen I think there is a meta thread about this topic...

looks at tb

I don't know what uniform boundedness principle is. But I want to make my own proof.

Not in the mood for reading.

Gigili

Does it make sense to you? "$g$ is measere "?

t.b.

@MattN The Banach-Steinhaus theorem

Pointwise bounded for a sequence of continuous functionals implies that it is uniformly bounded.

Matt N.

@Gigili No, I'm editing it. Hang in there.

Gigili

eh pft, I'm editing it too

Matt N.

2:25 PM

@tb Yes. I don't know what that is. Honestly, I can't learn from lectures or homework. Need to learn stuff on my own. And I haven't had time to look at that. Will do soon though.

@Gigili : )

@kahen Look, it's here : )

Américo Tavares

2:51 PM

@J.M. When you can give me some advice, please write it as a comment to the question or email me.

J. M.

I would more likely write it as a comment or an answer, so that other people can easily scrutinize my line of reasoning...

Matt N.

3:08 PM

Hello @Matt. Can you help me prove $\|f+g\|_{L^p (Z)} \leq \|f\|_{L^p (Z)} + \|g\|_{L^p (Z)}$? I've shown that $S = \{f \in L^p \mid \|f\| \leq 1\}$ is convex. Now I'm stuck.

Hm... let's see...

t.b.

Hint: Apply what you know to $\tilde{f} = f/\|f\|$ and $\tilde{g} = g / \|g\|$.

@MattN ...use that $\frac{\|f\|}{\|f\| + \|g\|} + \frac{\|g\|}{\|f\| + \|g\|} = 1$.

Matt N.

Hold on, just gimme a minute! : )

@tb Thank you!

t.b.

Got it?

(dumb question, probably)

Matt N.

@tb Yes.

t.b.

3:25 PM

So here's what I had in mind: for $0 \leq \lambda \leq 1$ you have $\|(1-\lambda) \tilde{f} + \lambda \tilde{g} \| \leq 1$ by convexity of $S$. Now put $\lambda = \frac{\|g\|}{\|f\|+\|g\|}$ and multiply through by $\|f\|+\|g\|$ to get the triangle inequality.

(so you were done already :))

Matt N.

Yes, I can see that. But I wonder how long I'd've had to think until I'd have thought of what to use for lambda. I was ~~wandering~~ wondering down a totally wrong path.

^puts on Italian accent

t.b.

:)

robjohn

@tb Not only a duplicate, but a special case.

Matt N.

But I'd like it to stay open, I like the question.

t.b.

@MattN It's one of those tricks I had to learn the hard way... These convexity arguments are always easy, but only after you found them :)

robjohn

3:30 PM

@MattN how is that question better than the other?

Matt N.

@robjohn That's not what I was saying. They're equally awesome. : )

robjohn

@MattN We are both talking about the Riesz Representation question, right?

Matt N.

Yes.

t.b.

@robjohn well, it's even a multi-dupe given the links in the question you answered. But on Matt's pleading I don't insist and wait for others to vote first.

robjohn

@tb I'm not rushing to vote on it. I just don't really understand (a common state of affairs :-)

t.b.

3:38 PM

@robjohn I was actually hesitating because it is a special case. If Matt wants to do this exercise and post his answer for us to review that's a good enough reason for me not to vote on it before others do.

Antonio Vargas

Could someone more polite than me give this question some help?

robjohn

@tb That's fine.

Matt N.

I was actually planning to do that. Caught me red-handed.

robjohn

I notice that the helpful flags are no longer part of the candidates statistics, or am I missing them?

@MattN Not hard to do with the deleted answer there :-)

t.b.

@robjohn they're still there for me. You got 4 of 'em

J. M.

3:42 PM

@robjohn I see "helpful flags" lines too.

robjohn

going back to look :-)

leo

hello :)

robjohn

I was looking at the wrong page :-)

t.b.

Hi, leo, how's it going?

Américo Tavares

@JM That's the best way, I agree. Thanks.

leo

3:47 PM

@tb enjoying a pretty nice sunny Sunday :)

I'm about to start the work

t.b.

@leo Good for you! I can send you some rain from here if you want :)

leo

@tb another dupe?

of this

t.b.

@leo yes, I just voted. A multiple multi-multi-dupe I guess.

leo

Let me count...

t.b.

@leo the duplicate you link to has five more questions linked. I'm not willing to climb up the dupe-tree further...

leo

3:56 PM

Yep. At least 5 related questions. Some of them with different dresses only

Gigili

@AntonioVargas I doubt there is such a person.

Does my answer make sense?

t.b.

@Gigili yes, if you explain the deep theorem that $n=x$ :)

Gigili

Umm, I just wanted to say it's really that easy.

It's going to be difficult that way, Tee-Bee!

Should I delete it?

t.b.

@Gigili no, why? I was merely saying that you could consider replacing $n$ by $x$ in the answer.

(it does make sense to me, in any case and it's a good heuristics to have in mind)

Jeff

@robjohn this is the last step with this prob. i'm going to work out everything myself. if you got a moment, can you confirm if gamma, in chat.stackexchange.com/transcript/message/4435823#4435823, was the whole curve in i.sstatic.net/mj8RA.png

Martin Sleziak

4:03 PM

It seems to be one of the abstract duplicates.

t.b.: I guess by n was meant the degree of the polynomial.

I just noticed.

@tb Ah yes, I will.

see you all :)

Bye!

Hello again! Advise me please some book where I can find the proof of the domain invariance theorem for manifolds

Jeff

4:07 PM

anyone know a quick way to type 'Res' as an operator in math mode in Latex?

Martin Sleziak

\operatorname{Res} ?

Jeff

@MartinSleziak ok. i guess i could have remembered that one! :D

Martin Sleziak

Or perhaps use newcommand to make you own macro for it.

Jeff

@MartinSleziak a macro is better. but i only need it a few times for now and am in kinda hurry (for a change!).

thanks

t.b.

@Nimza That's just linear algebra: if you know that there's a local diffeomorphism between two manifolds, consider the derivative in charts and use that there's no linear isomorphism $\mathbb{R}^n \to \mathbb{R}^m$ if $n \neq m$.

Nimza

4:12 PM

@tb thank you again :)

t.b.

The hard work in Brouwer's theorem on invariance of domain is in the extension from the smooth to the continuous case.

That's in every book on algebraic or differential topology I can think of: Hatcher, Bredon, Bott-Tu, Hirsch, Milnor, etc.

/me rushes to add Lee in order to make Jasper happy

Nimza

@tb thanks! And do you know, are there some sufficient conditions on function $f$ such that a continuous image $f(A)$ of the compact simply connected space $A$ is simply connected? I can't check the global injectivity of $f$, but it is locally injective

t.b.

@Nimza what is the function you're having in mind?

Nimza

@tb $x \mapsto x \circ \nabla g ( p \circ x)$ where $\circ$ is the entrywise product. here $x$ is from positive part of unit sphere. I know that Jacobian is nonzero in every point. $\nabla g$ is positively homogeneous of zero order and positive

t.b.

4:27 PM

Sorry, what's $p$? A fixed point of the sphere?

Nimza

some fixed positive vector. For example from the sphere but is doesn't matter since $\nabla g$ is positively homogeneous of zero order. "Prices" of "products" x :)

Gigili

Hä?

Noted.

Nimza

@tb my final target is to show that such $f$ is globally injective

Antonio Vargas

@Gigili That's kind, but I was tempted to send him to lmgtfy...

t.b.

@Nimza I understand but I don't quite see it right now. Sorry

Matt N.

4:35 PM

@tb And me! : )

Benjamin Lim

@tb Can you help me please theo

Gigili

Dupe: stackoverflow.com/questions/10471752/prove-fingerprinting

Haha?

Benjamin Lim

@MattN vector calculus stuff are you familiar? I have posted a question on the main site

my friends are having a lot of trouble with a vector calculus question

Matt N.

@BenjaminLim Not right now, sorry, I'm working.

Benjamin Lim

@MattN Ok thanks anyway :D

@JM Are you around??

@robjohn Are you around?

Antonio Vargas

4:53 PM

@BenjaminLim if the curves were traversed in the opposite direction then there would be a + on the LHS instead of a -

Paul Manta

Hello.

Antonio Vargas

Hello @PaulManta

Paul Manta

Is there any common symbol for "not necessarily equal to"? Using $\ne$ seems to suggest that two things are never equal.

Antonio Vargas

I've not necessarily seen one.

You could say something like "$x \ne y$ in general"

Paul Manta

@AntonioVargas That'll do, I guess.

Thanks!

t.b.

4:58 PM

@BenjaminLim what's up?

Benjamin Lim

@tb how's your vector calculus

@AntonioVargas what do you mean?

@AntonioVargas Both the curves are traversed in the anti- clockwise direction

Antonio Vargas

Indeed. So,

t.b.

I don't use it all that often. What's the question?

Benjamin Lim

@tb My friends are freaking out over some vec calc question

@AntonioVargas wait can you said what you said again?

I have some ellipse

it's inside a circle

both are traversed anti - clockwise @AntonioVargas

@tb math.stackexchange.com/questions/141830/…

Antonio Vargas

I know, typing. Give me a sec.

The boundary of $D$ with the natural orientation (in preparation of the use of Green's Theorem) is the union $(-C_1) \cup C_2$, where $C_1$ and $C_2$ are traversed in the counterclockwise direction. Writing $-C_1$ means "traverse $C_1$ in the opposite direction". So, Green's theorem gives us... (typing)

Benjamin Lim

5:04 PM

@AntonioVargas Can you type up an answer?

Antonio Vargas

Sure, no problem. That's easier than chat.

Benjamin Lim

I may upvote and accept your answer

Antonio Vargas

nulluser's answer is correct, do you still want me to post mine?

t.b.

geee. so what's the hurry? you asked the question 20 minutes ago and you ask for help as if the world's on fire :)

Benjamin Lim

@AntonioVargas Look at the comment I posted on his answer

@tb I don't do the course

Dylan Moreland

5:07 PM

All the more reason to be at peace.

Benjamin Lim

@DylanMoreland I just realised that I can use direct limits to show that the torsion submodule is isomorphic to some tensor product

@AntonioVargas how is the answer coming?

t.b.

@DylanMoreland I suspect it's the girlfriend :)

Antonio Vargas

@BenjaminLim Fine, keep your pants on :)

Benjamin Lim

@tb I don't do vector calculus, my vector calculus is so rusty that I don't even remember the statement of green's theorem. My friends come to me for help at 2.30am on their assignment, this is why I am on Math.SE now!

so tired

Antonio Vargas

@BenjaminLim, just posted it. Is it clear?

Benjamin Lim

5:16 PM

@AntonioVargas Ok that's clear

my friend figured it out :D :D :D

Antonio Vargas

......

t.b.

heh :)

Benjamin Lim

@tb I ask questions on galois theory and commutative algebra all of a sudden vec calc?

Dylan Moreland

I don't think he's doubting you.

Benjamin Lim

@AntonioVargas Why would I all of a sudden ask about vec calc?

@DylanMoreland I have gone over to the dark side of the force.

t.b.

5:20 PM

keep cool, Ben, no need to defend yourself...

Benjamin Lim

@tb hahahahahaahahahahahahah :D :D :D :D :D :D: D: D: D: D: D: :D :D :D :D

@tb Just curious, but why do keyhole contours work?

N3buchadnezzar

Is not the point that you eventually make the keyhole infitesmal, so you are basically looking at a continous portion of the real axis ?

Benjamin Lim

@N3buchadnezzar I don't know I've never done this sort of stuff in my life

t.b.

one of those marvelous edit histories...

N3buchadnezzar

@BenjaminLim My understanding of it is that an integral in the complex plane always is a line integral, or a path integral over some closed curve.

Benjamin Lim

5:29 PM

@N3buchadnezzar ok

@tb That is ridiculous

Antonio Vargas

@tb I don't see Michael's goal there...

t.b.

@AntonioVargas Let me put it this way: he's uhurm... slightly obsessed with The Right Way to Use TeX

Matt N.

@tb What if I edit them back in?

Benjamin Lim

@tb Who is Michael Hardy?

Matt N.

Ah, not funny. He'd just roll back...

N3buchadnezzar

5:31 PM

I think it is fine, I am just glad he is not my mum!

Antonio Vargas

@tb I seee.

N3buchadnezzar

I do not want to eat dinner...

robjohn

@Jeff The curved part really goes to infinity (or a close approximation)

@BenjaminLim I am here now (took a short nap)

Benjamin Lim

@robjohn Ah it's ok now

N3buchadnezzar

@robjohn Do they have to? I have seen contour integration being used on integrals that are not over the whole real axis.

Benjamin Lim

5:33 PM

I had to help my friends with some vector calc

N3buchadnezzar

Bear in mind, I have little knowledge in this field for now.

robjohn

@BenjaminLim Sorry I missed it :-)

Benjamin Lim

@robjohn It's ok

Antonio Vargas

A picture of calculations on a brown paper towel?

Benjamin Lim

they figured it out

they are all freaking out

t.b.

5:34 PM

@MattN That'd be funny but I don't want to risk a heart attack on MH's part :)

robjohn

@AntonioVargas Physics should be kept in a brown paper wrapper?

Benjamin Lim

@tb Who is MH?

robjohn

Mike Hardy

Benjamin Lim

@robjohn I meant rather where is this guy now

N3buchadnezzar

I would love to edit his post using mathtype latex, conversion.

J. M.

5:37 PM

Let's not be too mean... :)

robjohn

@N3buchadnezzar In the problem given, yes, but contour integrals that surround the same residues are equal (that's the whole point).

t.b.

But the funniest part is that he manages to produce crap like this

N3buchadnezzar

@robjohn Indeed, then it is just a matter of picking a contour that lead to nice symmetries.

robjohn

@N3buchadnezzar Well, contours that give you what you want and you can compute :-)

Benjamin Lim

@tb I have edited that

N3buchadnezzar

5:40 PM

I would then preffer starfish contours, or yellyfish shaped contours.

Matt N.

lol

J. M.

@N3buchadnezzar You can do that, but the algebra will likely be horrific...

robjohn

@N3buchadnezzar In the problem given, we want to compute an integral over half the real line, so the bottom part of the countour should be essentially half the real line.

N3buchadnezzar

Indeed. I guess the worse possible contour you could do would be half of a koch snoflake.. Where the bottom is flat.