I'm a bit worried about Marvis going on a rampage because of what he feels is an unjustified downvote. Last time he burned his account because of some not-so-nice exchanges like that.
I remember a feature that I could see the number of downvotes and upvotes by clicking on the number next to the question or answer, is it a 3k user thingy?
Do we have a post at meta explaining how to vote? That buttons called 1st/2nd/3rd choice seem to be self-explaining; but maybe it would be interesting to know: Is it possible to change my vote, if I decide to? Can I see somewhere how many votes the candidates already have?
@PeterTamaroff I did. I interpreted your insistence as serious enough, but I recompensated you with two upvotes. It was impossible to give you badges alone.
@BrianMScott Are you speaking about yourself? (Well, you're gravatar is green, I have to admit you're a bright person; but why stating the obvious...) ;-)
@PeterTamaroff so, you've got two shiny badges instead of two points. Sounds like a fair deal, no? When I was having 33335 points no-one was willing to give me the requested downvotes, they upvoted me instead :)
@JM I only found this post, where someone is asking whether he has to use all of his 3 votes: Question on election voting. I did not find much more explaining details of voting process at meta.MSE.
@MartinSleziak I forgot which thread, but there was somebody who said that he did "first choice on one day, and the other two choices on the last day".
I was under the impression that $u(z)\colon\mathbb{C}\setminus\{0\}\to\mathbb{R}$ defined by $u(z)=\ln(|z|^2)$ does not have a harmonic conjugate $v\colon\mathbb{C}\setminus\{0\}\to\mathbb{R}$.
If $v$ is a harmonic conjugate, then $f=u+iv$ is holomorphic. However, $\frac{\partial}{\partial\theta}f(re^{i\theta})=ri\frac{\partial}{\partial r}f(re^{i\theta})$. So $$\frac{d}{d\theta}v(re^{i\theta})=r\frac{\partial}{\partial r}u(re^{i\theta})=r(\ln(r^2))'=2.$$ But then integrating over $\theta\in[0,2\pi]$ yields $0=v(re^{2\pi i})-v(r)=4\pi$, a contradiction. Is there something wrong with this argument?
@JM Well, that's about the only thing that was really hilarious in the German synchronized version of the Flying Circus. They translated the charades verbatim. So, for example apple, coffee --- ah, not guilty! huh?
@PeterTamaroff Charades are a game where you have to guess words. Someone gets a word he must describe by acting but without speaking and others guess it. Here's the Spanish WP page explaining it.
@PeterTamaroff Nothing, as far as English speakers are concerned: most would, I think, be vaguely aware of some relationship with the word mime, but pantomime stands on its own as an independent word, borrowed from Latin. You have to go back to the Greek source to see the elements from which it was originally compounded.
Oh, I almost forgot: in Greek the first element of the compound means 'all'; I don't remember offhand why.
@tb It's not actually clear. Greek mimos could mean either 'a mimic' or 'a mime' (in the 'actor' sense), and the compound is apparently relatively late $-$ Hellenistic $-$ so I'm not sure exactly what the panto- was understood to add.
@BrianMScott It was a shot in the dark that seemed to make sense, but closer scrutiny often proves the contrary in those cases. Thanks for going after it!
@Gigili, to inflame, among other things, means *to provoke (someone) to strong feelings.* nowadays this usage is out of fashion—it was very popular among romantic poets in the early 19th century. :)
@MarianoSuárezAlvarez I see. In fact I looked it up in the dictionary to avoid further misusing! I was unsure which comment of mine are you referring to.
No, we have 0M+a, 1M+a, 2M+a, ...., qM+a as the elements of $S_a$.
And it's only one equivalence class, that of $a$ modulo $M$, intersected with $[N]$.
This is only if a is less than or equal to r so that qM+a<=qM+r=N, ie the last element I listed is in range. If not, you have to chop off the last element and there are only q.
@JM, I just stumbled on the WTFPL when I read your profile. Cool stuff ! I'm going to use it more now. I was hoping something like that was already in creation. Thanks!
Sorry, but it was given for 0≤a≤M−1, S_a... Its a bit confusing for me. Could you please explain with an example, say N=5 and M=3. I've only primary knowledge about equivalence classes.
Okay. List out all numbers congruent to 2 modulo 10 that are less than 55. Now list out all numbers congruent to 7 modulo 10 that are less than 55. Here we have N=55=5(10)+5, so that M=10 and r=5. The first has a=2<r and the second a=7>r.
nonnegative numbers, if you want to be persnickety
@Nunoxic No worries. If I were concerned about my "methods" being used by other people, I wouldn't post them here in the first place, so that one seemed appropriate.
@Andrew You have to use formulas first to reduce the problem. In other words, write down a formula for |Sa|, then sum over all 0<=a<=M-1. (And discard the 0+0 solution, eh.)
You'll need to split the sum into 0<=a<=r and a>r.
Being a contest problem, I think I've explained enough at this point. :)
Ah yes, a=b is another case that needs to be undone in the calculation. Yes on the b for each Sa, but you need to take out the symmetric solutions with b>a.
I'll talk about that last point. Remember when I listed out the elements of Sa? That last element was qM+a. If a>r, then qM+a>qM+r=N which is outside of range, and you can't count that one.
@Nimza: just say "the topology of uniform convergence of $f_n$ and all of their derivatives on compact subsets of $E$". That's much easier to type and mentally parse.
There are criteria on when a functional on $C^\infty$ can be represented by functions or nice measures.
This should be in every book that treats distributions seriously, but I can't reproduce these results right now.
But in general they cannot. For example, you have the nice functional $f \mapsto f'(x_0)$. No way that this can be represented by a function or a measure.
@tb Sure, I was only half-serious : ) (as almost always)
I'm sitting in the class room and the marketing people are still in there (they have the lecture before ours) and they're discussing what grade the student giving the presentation will get. Now the "lecturer" told him "We've been discussing whether we will give you a 5.75 or a 6.0. Usually we don't give a 6.0 but..."
It makes a lot of sense, you know :) You've got 3 negative grades and two positive ones, so let's remove .25 from the positive grade scale, what the hell? Basically they could introduce a scale: one grade for passed and a very fine-graded scheme between failed and flunked abysmally, best measured by grades precise to the .001 points. Especially if you have ten criteria of which 12 are subjective.
You lost me somewhere in the middle of the paragraph : D Need to read it again.
But eeww. This is so low. It's comparable to feeling like a man and all strong and whatnot when having gun while everyone else is unarmed. Or when shooting bears, foxes and whatnot. Very manly. Wooow.
@tb Haha : D But on a more serious note: Why not have 1 and 2 where 1 means failed and 2 means pass? I think I'd thrive in a system like that.
@MattN yeah, why not. I don't understand what grades precise to the .25 between 4 and 5.75 are supposed to mean, especially when it comes to judging a presentation. So you've got 8 points between which you can choose. What if I think that I can only judge precise to the .5 grades and don't give grades above 5, because no-one's better than good?
@tb Well, I wasn't trying to say I was. But by giving 1 or 6 I do two birds with one stone: I please the students and I make a statement about the grading system namely that it's unnecessarily accurate.
