@Venus finally found a book that treats a bit on asymptotic and some probabilistic methods see: www.ams.org/bookstore/pspdf/stml-71-toc.pdf but i can't find it online : ((
There is another way to do it.
You first divide the $n-k$ non-selected positions into $g+1$ blocks so that the $g-1$ middle blocks have at least one non-selected item each.
By stars and bars there are $\binom{n-k+1}{g}$ ways to do this, after this separate the $k$ selected items into $\binom{k-...
DanielFisher deserves three upvotes on each of his answers, the first one for his sympathy, the second one for his mastery of the subject, and the third one because he's too damn awesome!
@DanielFischer is there something like the function in facebook where you can see your interaction with another user? Like look at questions in which you both commented or comments on your posts by that user or visceversa? I think that would be uber-cool. @PedroTamaroff
@DonLarynx gonna group-theory the shit out of problem 281
@JorgeFernández I think there is such a query on SEDE, but I'm not sure. It's not something the Stack Exchange sites were made for (on the contrary, rather).
@MikeMiller Yes, continuous extension to the boundary is a local issue; its existence depends on the structure of the boundary near a particular point. The other components don't matter.
One can derive this from the simply-connected version of the same fact. Fill in all holes except one, then map to a disk. Restore the holes. Now you have a conformal image of the same domain, but one of the boundary components is an analytic curve. Repeat for each boundary component. You get a domain bounded by analytic curves.
I posted here that I didn't know whether it had an extension to multiply connected domains. If you care to comment with your insight I'd appreciate it.
This was the proof I tried to write down, but the issue is that the original conformal map needn't extend to a conformal map of the filled-in region.
e.g. pick an annulus of the right radii, and take the map z \to 1/z.
@MikeMiller But the question does not seem to be asking about continuity. It's just "boundary goes to boundary". That's true for any homeomorphism of open sets.
I made the assumption that it extends continuously to the boundary, because otherwise it's difficult to interpret what it should mean that boundary goes to boundary.
So the interesting question is when you can extend thus.
Please clarify what you mean by "take boundaries to boundaries". Presumably, it's a statement involving some limits; make it precise. — Behaviour17 secs ago
@MikeMiller Since you're at UCLA, you should have Harmonic Measure by Garnett and Marshall nearby. There is a chapter on finitely connected domains there, and I'm pretty sure the result can be found in that chapter.
@DonLarynx I wanted to start studying in 2015, but my mind got really messed up recently, and I am not sure why that happened. So right now, I won't be studying. I will be planning what to do for the rest of my life. I will see what changes I need to make in my life.
@user153330 Yes, I understand that. But it's really more complicated than that.
@JasperLoy that's normal, that's actually because you were near at starting to make something really big, but at the very last moment you had some doubts. all i can say is throw those stupid doubts out the window and start
@quid Kofia is "first post on a site between 12/26 and 12/31". My first post on MSE was long ago, and thus I am disqualified. I got it for a post on Arqade.
@user153330 I have not been working for 7 years. This period, I did my best to get well and get on with life. But lots of shit happened these 7 years too, and often things got worse. The future seems hopeless. But I am hanging in there and trying to create some miracles with what little I have left.
we wish to show that $PO(1, n-1)$ is transitive is on this set. This is easy for time-like vector subspaces. Why, and what is the main difficulty when the subspaces are space-like?
@Chris'ssis: No. Most of my friends are on holiday in the mountains or so with their families, and the friends that are here like to go clubbing, but on New Years that's a really bad idea because there are way more people and everything is way more expensive.
@Chris'ssis: I'll just watch the epic fireworks through my window, I suppose.
@DanielFischer On a square contour centered at the origin, how far away from the positive integers would the contour need to stay so that when $z$ is large in magnitude, $\psi(-z) \sim \log(-z)$, $\psi_{1}(-z) \sim -\frac{1}{z}$, $\psi_{2}(-z) \sim - \frac{1}{z^{2}}$, etc.? Can you get too close? I've never really thought about this before.
@RandomVariable Never really worked with them (well, one uses $\psi_2$ to show a couple of things about $\Gamma$). Perhaps you can get useful estimates by playing with the series representations. But ask robjohn, he knows the polygammas, I believe.
I wonder how these people manage to go on a ship at this time at this date. I was on a ship once in autumn and already found it really cold, and now it's 30° colder.
@quid: There's a ship on the lake here which goes for a short tour around the lake (~2 hours) till late in the night. And I just saw it, but I didn't know it was still going and there were a lot of people on it.
@DanielFischer I'll ask him. It's an issue that comes up when using contour integration to evaluate sums that involve the harmonic numbers and/or generalized harmonic numbers. I've posted a few answers using that approach. I'm not sure if those answers need to be edited a bit or not.
Has to be after UTC new year. To-do list for New Year celebration: (1) post in chat; (2) enter an illegal star-exchange agreement; (3) delete 10 comments.
It's much beauty in there, a world of amazing connections! Just think about it: you solve an integral that at first sight it seems you won't ever use it elsewhere, and then, after a while, you realize that modifying that weird integral in a certain way, you manage to evalute integrals that are not known to have beed evaluated before.
@AndrewThompson I think people tend to be good at what they like. If I did something I do not like I couldn't possibly get good results. Maybe other would do, but not me.
@Chris'ssis Probably a perfectly valid point of view, but predetermined aptitude is a factor.
@Chris'ssis I often fear that I am clever enough to appreciate mathematics but to dull to contribute to it, which may or may not be true. Ask me in a couple of years.
There is another way to do it.
You first divide the $n-k$ non-selected positions into $g+1$ blocks so that the $g-1$ middle blocks have at least one non-selected item each.
By stars and bars there are $\binom{n-k+1}{g}$ ways to do this, after this separate the $k$ selected items into $\binom{k-...
