@robjohn Something someone said in chat was flagged offensive, and you have 10k so you can see the flag and vote on it. Mods handle them pretty fast though, that's why the circle goes away
That's right. It's also much more relaxed than going off-topic in the comments. Also, I like it that some users pop in and ask some easy questions. For some things it's much better to do this interactively than in the more rigid Q&A format.
And since robjohn adapted this script, it's really comfortable talking math!
Not sure how deserved it is. The first 50 votes or so came in before the long addition and discussion, where the answer just said "this is the liar paradox, nyah nyah, it's meaningless".
Oh, today an answer I posted over half a year ago was accepted. I couldn't believe it. But it's one of my personal favorites. I had a lot of fun drawing with Geogebra...
@tb I came up with what appears to be a new result related to Soddy's theorem. I talked to several people who are active in that area, and they did not know it before.
Not much of a signal no. But if the DSL modem buffers up to, say 5 MB of outgoing data and doesn't shrink that buffer when the uplink is slow, that means that it can saturate with several seconds worth of data without your computer knowing.
Intuitively, one would think providing a larger buffer instead of just throwing packets away would improve performance, but it can really bomb the TCP/IP suite to have too much buffer.
@robjohn It's the most likely explanation why I couldn't even connect to the server from outside. Under better conditions, even with harshly limited bandwidth, the TCP protocol would have shared equally between me and the other user instead.
(Also, I converted bits to bytes the wrong way. You don't need 5 MB of buffer to get into trouble at 300 kbps -- just a few hundred KB could wreak havoc).
10:1 is good enough for back-of-the envelope estimates, but I don't think start and stop bits have actually been involved in the quoted speeds since the days of dialup links and serial cables.
That's what I suspected. Number of nodes, right? So may it simply be a shift where one starts counting? I mean your formula reads $$n_0 = \sum_{j=3}^{k} (j-1)n_j + n_2+1,$$ right?
(I'm not very familiar with graph theory, but it seems extremely close to what's written there)
Oh, that one (=Gigili's question) is a bit cute. =) I am not fully sure, but I think the point is this: the degree of any node equals the number of children plus 1... Except that this does not work for the root.
If X is limit point compact and f: X -> Y is continuous, does it follow that Y is limit point compact? I have a "proof" the answer is yes, but apparently a counterexample exists. Does anyone know what it is?
Consider $f(p)$. If this were not a limit point, there would be some neighborhood in $f(X)$ of $f(p)$ that is empty except for $f(p)$. Call this neighborhood $N$.
Look at $f^{-1}(N)$. This is a neighborhood of $p$.
So it contains a point $j\in U$, because $p$ is a limit point. Then $f(j)\in N$, a contradiction.
The counter example is this: Consider the set $\{0,1\}$ under the indiscrete topology. Then the projection from $\{0,1\}\times \mathbb{R}$ to $\mathbb{R}$ is continuous. The domain is limit point compact, but the image is obviously not.
@ZeeshanMahmud So, just for the stupid like me to understand: after top people compiled a list of gigantic challenges that are as widely open as can be you want to compile a CW list containing even greater challenges by asking some random people that happen to be interested in all kinds of rather basic math?
@DylanMoreland I guess a lot of things can be done with conics without resorting to calculus. Most of it is either forgotten or ignored or generalised to the point of no return =)
@ZeeshanMahmud And remember what the FAQ says: You should only ask practical, answerable questions based on actual problems that you face. Chatty, open-ended questions diminish the usefulness of our site and push other questions off the front page.
@JM Thanks, JM. I am not completely sure of how things work for cubics (although I can imagine doing the same thing, except that the discriminant bit gets replaced by a longer expression).
@JM It's quite sad, you know. I learned about these conics while preparing for my entrance exams. There the emphasis was on getting to the answer as quickly as possible, so most of the times, we just ended up remembering a hideously long list of formulas (I don't even remember what these are anymore). I don't quite know the "theory" behind these things. :-(
@Gigili [To be frank, I am still thinking about the other answer...] I don't mind you posting it as a question. Or, I can compose an answer offline when I am in the right mood, and ping you later. Either works for me.
@JM Would you happen to know any good treatment of this stuff? I am looking for the "pull up the sleeve and do the algebra" kind.
@Srivatsan Well, to add to that. Gigili, you can simply say: I'm interested in binary trees and I'm unable to translate the answers given in the other thread into my terms and conventions. Can you lend me a hand? I suspect something along the lines ... And I'm pretty sure people won't vote to close that.
@Srivatsan Ah, I managed to remember one of them. If you're well-versed in algebraic geometry already, that's a plus, but it's not a necessary prerequisite.
J.M. has given a very good answer explaining singular values and how they're used in low rank approximations of images. However, a few pictures always go a long way in appreciating and understanding these concepts. Here is an example from one of my presentations from I don't know when, but is exa...
@JM Typically, they're all treated as three separate matrices and operations act on each of them. I chose to convert to grayscale here, since it's simpler
Aw, I now see what Greg was talking about. Sort-of...
Fix the map $a_i \mapsto a_{i+k}$. Under this map, we get the cycle, $a_0 \mapsto a_k \mapsto a_{2k} \mapsto a_{3k} \mapsto \cdots$. So we need to show that each $a_j$ appears in this list somewhere.
