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12:27 AM
I think it's time to go to sleep.
Thanks for today, @Srivatsan, @tb, @robjohn.
Good night.
 
12:42 AM
hi
 
 
3 hours later…
3:31 AM
@JM Reg. math.stackexchange.com/questions/91325/…, I don't mind if the second commenter posts that as an answer. I am deleting my comment... =)
Ok never mind. I won't delete my comment since your comment as well as Gerry's are based on mine.
 
@Sri: I see your point and Gerry's so it's fine with me. :) No need to feed a schmuck, after all...
@Srivatsan What a complete arse...
 
@JM You saw that last comment, right? =)
Yep, complete arse seems apt.
Personally, I haven't got my finger burnt because of this user, but they really shouldn't do this to anyone.
I don't see why we are still putting up with the complete lack of cooperation and the teasing.
 
@Srivatsan Unfortunately, this doesn't look to be one of the things we can readily take to meta, lest we be seen as bitching. Maybe we should just all ignore him, downvote his crap, and hope he gets the message...
 
@JM Hm, ok. I am obviously on the side of the ignoring -- as that comment suggests.. =)
 
@Srivatsan Last I heard, even the font's creator was genuinely repentant for inventing it...
@Srivatsan Sometimes, it sucks that you can't punch people in the mouth on the Internet...
 
@JM I am quite pissed at this. I think I will just take a stroll and then go home...
 
@Srivatsan That might be good. Enjoy. :)
 
@JM Thanks. See you later.
And good morning.
 
Hi and bye Srivatsan
 
4:00 AM
Hey t.b.
 
Hey, J. M.
That was a short transcript...
 
hi tb
 
@tb Yes, it's odd how short it is... :D
 
I'm surprised that this thread is still open and while I'm at it how did this ram it down their throats answer get 9 upvotes?
 
@tb It still needs an executioner. I was the last juror.
 
4:04 AM
Done. =)
 
But what's the point of this "ram it down their throats" non-answer? I must be missing something.
 
It's an outlet for frustration?
 
I can't see anything redeeming either.
 
And after reading about the sum of cubes, I'm frustrated enough to make irrational upvotes...
 
What's up with the sum of cubes?
 
4:08 AM
@tb Don't tell me: you haven't seen this question: math.stackexchange.com/questions/91325/…?
 
Sri linked to it 12 minutes ago. Sorry I'm mobile or I'd copy it
 
@tb OP is being a complete arse, as usual.
 
I rolled back. This is vandalism...
 
@TheChaz Oh, I haven't read the question properly. It is an intriguing possibility to indulge in your self-infatuation... Never occurred to me :)
 
4:12 AM
(hearty chuckle)
 
I love it when an answer starts with "Hint I think..."
 
Re: sum of cubes, where is the unilateral deletion of comments by Jeff Atwood when we need it?!? :o
 
@TheChaz Jeff hasn't visited meta in a while... :)
 
@TheChaz I know you're joking, but seriously: We don't need it. I would rather let the comments stay, as an evidence how much of a jerk the OP is.
 
I wish we were all moderators!
I've gotten used to taking unilateral action against jerks and spammers on another math site (forum), so it's hard to see garbage fester.
 
4:21 AM
@Srivatsan There was also this question
 
@tb Yes, I flagged for deletion.
Like stabbing a guy that is dying of a heart attack anyway. =)
[But heh, I got my 10 flag weight. :=)]
 
Yeah, you're getting 50 times more per flag than I do...
 
Isn't this distinctly nonanswer-y?
@tb Wait, that's not possible. You get less than a point per flag?
 
@Srivatsan Yes, we do.
 
@Srivatsan yep, at the moment I'm getting around 0.2
 
4:26 AM
Past 500 (IIRC), the increase is not as great as it was.
 
@JM That sucks very much. How do you guys manage to gather so many points then?
 
Asaf was particularly pissed about how the weight increase was tiny, but the penalty for a declined flag was high.
 
I say: stop flagging posts yourself, and give me a chance. =)
@JM - what's the penalty? 2 points?
 
@Srivatsan -10.
 
@JM Ah, that's too high indeed.
Wait, 50 helpful flags will offset one mistake?
 
4:29 AM
@Srivatsan It remains -10 per invalid flag, but the points you get per valid flag decrease exponentially. See here. I estimate that you need 500 valid flags to get to the max of 749.
 
@tb Oh, you're quite close to the ceiling then.
 
There are some explanations for this somewhere on meta.SO, but I'm too lazy to look for them.
 
Please don't inconvenience yourself too much.
 
@Srivatsan Precisely why Asaf soured on flagging after his last declined flag...
 
@Srivatsan Here's Asaf's rant or should I say whine?
 
4:33 AM
I'd say both, but I do agree with him.
 
I think Grace linked to the relevant meta.SO threads somewhere in that thread. ♪
 
Can somebody give a hint how to prove that metric $d:X\times X\rightarrow\mathbb{R}$ on the metric-space $(X,d)$ is continuous?
 
@RamanaVenkata what metric do you put on the product? It doesn't matter, but let's pretend it does.
I just noticed: my rep at the moment reads ten-nine-eight-seven.
 
Nice.
Ramana: that was one nice answer you gave here :)
 
@RamanaVenkata Ramana, if you didn't understand what on earth my question meant, then just say so. It's not too relevant.
 
4:43 AM
@tb Thank you...
 
Are there missing symbols in this answer of mine? The OP claims that it's garbled but I don't think I've done anything fancy.
 
@Srivatsan The metric function has two co-ordinates so I think $d:X\times X\rightarrow \mathbb{R}$ makes sense to me but what definition of continuity should I use is the question I am facing.
 
@DylanMoreland Looks ok to me, Dylan. No obvious problems.
 
@DylanMoreland Looks perfectly fine to me. Tell Manos to shift-refresh and if that doesn't work to empty the browser cache and restart the browser.
 
@RamanaVenkata Ok, just a second. I am editing your answer.
 
4:47 AM
@RamanaVenkata I'd use the characterization by sequences.
 
Thanks for the sanity check. Maybe I'll attach a screenshot.
 
You have $(x_n, y_n) \to (x,y)$ if and only if $d(x_n,x) \to 0$ and $d(y_n,y) \to 0$. Can you take it from here?
 
@tb Can't we use the $\epsilon-\delta$ definition?
 
Sure you can, but then you need to be specific what metric you put on $X \times X$. That's what Srivatsan was getting at.
I'd use the metric $d_{X \times X} ((x,y), (z,w)) = \max{\{d(x,z),d(y,w)\}}$ to be specific.
 
In mathematics, the product metric is a definition of metric on the Cartesian product of two metric spaces. As described below, the p product metric of the Cartesian product of n metric spaces is the p norm of the n-vector of the norms of the n subspaces: :d_p(\mathbf{x}_1,\dots,\mathbf{x}_n) = \|(d_1(\mathbf{x}_1), \dots, d_n(\mathbf{x}_n))\|_p Definition Let (X, d_{X}) and (Y, d_{Y}) be metric spaces and let 1 \leq p \leq + \infty. Define the p-product metric d_{p} on X \times Y by :d_{p} \left( (x_{1}, y_{1}) , (x_{2}, y_{2}) \right) := \left( d_{X} (x_{1}, x_{2})^{p} + d_{Y} (y_{1}, y...
 
4:53 AM
That would be the metric $d_{\infty}$ in Wikipedia notation.
 
In terms of the topology, you'll get the product topology on X x X. So you could go that route.
 
The problem is from Munkres book on Topology It is as follows
Let $X$ is a metric space with metric $d$. Show that $d: X\times X \rightarrow \mathbb{R}$ is continuous.
I think here the metric $d$ can be anything
 
Well, he would've defined what topology he assumes on the product metric.
I usually take it to be the $d_1$ metric (in wikipedia notation), but $d_{\infty}$ or $d_2$ would work as well.
 
@Srivatsan Okay.....
 
@RamanaVenkata Can you tell us what Munkres uses?
 
5:08 AM
I don't know I read the chapter but I didn't understand what he wrote...
 
:) NARQ
 
t.b. always has great ideas... :D
 
Voted.
 
@Srivatsan btw. V. would accept your answer on the sums of cubes thread.
(at least that's the way I read the last comment)
 
5:18 AM
@tb Well, but the OP is not ready to look at the previous questions? I am not too keen on posting it as an answer yet...
 
Sure, I just wanted to mention it.
 
Um, I was just reading the subsequent comments in that page. I am not completely sure what the OP means by "I accept these answers" -- it could be what you just said.
 
5:29 AM
@tb I think Continuous in each variable separately doesn't mean the function is continuous..
 
@RamanaVenkata I have never said that, but you're right.
I said $(x_n,y_n) \to (x,y)$ in the product topology if and only if $d(x_n,x) \to 0$ and $d(y_n,y) \to 0$.
 
Okay.. Small misinterpretation.
 
@RamanaVenkata are you still here?
We could go through the argument if you want.
 
Yeah tell me
 
Is it clear to you that the metric $d_{X \times X} ((x,y), (z,w)) = \max{\{d(x,z),d(y,w)\}}$ gives the product topology on $X \times X$?
 
5:43 AM
yes
 
So: we want to show that if $d_{X \times X} ((x,y), (z,w))$ is small then $|d(x,y)-d(z,w)|$ is also small, right?
 
Now write $$|d(x,y) - d(z,w)| \leq |d(x,y) - d(z,y)| + |d(z,y)-d(z,w)|$$ (this is just the triangle inequality for $|\cdot|$).
But $|d(x,y) - d(z,y)| \leq d(x,z)$ and $|d(z,y)-d(z,w)| \leq d(y,w)$.
So $$|d(x,y) - d(z,w)| \leq d(x,z) + d(y,w) \leq 2 d_{X \times X} ((x,y),(z,w))$$
And this gives you that you can take $\delta = \varepsilon/2$.
 
Okay...
 
 
1 hour later…
6:56 AM
Quick question: I want to show that the sum from 1 to infinity of log(1+i/n) diverges. The real part converges, so how do I show the imaginary part diverges? I think I'm missing something simple.
 
I guess the idea is that $\log \left(1 + \frac{\mathrm{i}}{n} \right) \sim \frac{\mathrm{i}}{n}$, so the series looks like $\mathrm{i}$ times the harmonic series.
 
Ah, right
Thank you
 
You're welcome.
 
7:18 AM
When iterating through Markov chain states (successive transition matrices), how do we know if a state is terminal? What is the mathematical test for a terminal state?
Of course, we can tell by looking and natural intuition. I'm asking if there's a formal way of stating it?
For example, probabilities sometimes will converge to zero, or tend to one as the number of "random walks" or "trials" increase. How do we mathematically infer if terminality has been reached.
 
7:45 AM
Top of the morrow to you, and whatnot.
 
anyone?
:(
 
@IntelligentMoron The guy who likes dealing with stochastics currently isn't around, sorry. :( You might want to try asking on the main site.
 
Howdy J.M. how do you do?
 
I think this thread has served its purpose voted to delete it.
Hi Asaf greetings and whatnot to you
 
8:01 AM
@AsafKaragila Feeling quite dandy, thank you.
 
The thread is now deleted. The execution squad has spoken!
 
I'm opposed to deleting the first one. Sivaram's answer's nice and Victor should suffer from some of his non-questions.
(I'm pretty sure the first vote to delete the thread is his)
 
We can undelete the ones we removed to make him suffer some more ;-)
 
Nah, that would amount to leaving spam on the site.
 
Huh... the latter took 4 votes to delete.
 
8:09 AM
Agreed, let's not leave too many broken windows...
 
But how will the breeze come in?? :-)
 
After deleting the second question I took the opportunity to get rid of the ridiculous tag
 
@AsafKaragila That is a very interesting bug...
 
@JM I know.
 
@JM Guess we hit the okay button simultaneously
 
8:11 AM
UGH! For crying out loud. They are striking again. Now the class of Sunday will suffer of being two weeks behind! >:(
 
Who's on strike?
 
The union.
 
...and since you aren't part of it, you think their demands aren't reasonable?
 
I'm glad this got closed. "I want advice, but I'm completely unwilling to tell you the particulars."
 
I can't quite understand how people can manage to ask for help, and yet be coy...
 
8:17 AM
@JM No, I am a part of the union - but I hate it when the strikes interfere with my work.
 
@JM to be blunt: the (probably justified) fear of being told "well, but that's just ..."
 
@AsafKaragila Ah, I now see your position. :)
 
@JM Heck if their demands were to be accepted and instated within this year - my paycheck will be doubled.
 
Sounds like a probable outcome, then, no?
 
@tb I see that, but it bugs me that most people don't seem to realize that bringing out new stuff's always a gamble...
 
8:20 AM
@tb I doubt that. There's a little chance for that specific demand to be accepted; and even if it will be it is even less likely to be instated this year. Next year I'll be a Ph.D. student and my paycheck will be doubled anyway.
 
I was trying to say what you're saying in the first sentence...
 
Oh that.
Although this specific demand is quite reasonable.
Many M.Sc. students working as lab assistants, homework graders, or have office hours and that's it. They required that M.Sc. students actually teaching will get paid more.
 
They aren't doing that already?!
 
@JM I get more work hours, so I get more money. But my hourly wage is the same as any M.Sc. student working in the university.
 
Let me get this: you're teaching and someone else is correcting?
 
8:27 AM
Huh?
 
homework graders: what do they do?
 
Wait, wait... Asaf, you're a TA, but not actually teaching, no?
 
Grade homework assignments.
@JM Four frontal hours every week.
Furthermore, due to some issues of all teachers in the course - it ends up that I actually teach them some of the material as well.
 
@AsafKaragila I'd say that's a nastier problem than mere wages...
 
@JM It's fine for now, since I'm TA'ing logic and set theory. I'd much rather teach though.
 
8:30 AM
I'm used to the system that TAs are discussing homework assignments. I'm not quite sure how you can teach people what they need to know and what they did wrong in their solutions if you haven't corrected the assignments. That's why I'm asking.
 
@tb In here it really depends on the level of the course. Next semester I'll pretty much discuss integrals from homework sheets for four hours a week. This semester I'm teaching the top theoretical course (which has an exercise class) so I review definitions and explain them.
 
Good morning everyone.
 
Anyway. We have a seminar in 30 minutes and I have yet to shower. I shall be gone now.
 
and someone else corrects the actual assignments?
I don't get it.
Morning, Matt
 
@tb Undergrads (sophomores and such)
 
8:32 AM
Hey Matt.
 
I used to do that when I was in my undergrad. I hated every moment.
 
Hey all
 
@AsafKaragila Now I'm hopelessly confused.
 
Hey @Matt
 
@AsafKaragila Sounds like a very convoluted system...
 
8:33 AM
Nah, it's not. I'm just rushing to leave so I can't sit and explain ;-)
I'll explain some other time.
See you guys later.
 
See you!
 
Have fun Asaf!
 
Hey Matt. I was offline for most of yesterday (morning), so couldn't help with the uniform convergence thing. Anyway, how did it go?
 
It's ok, I figured it out I think.
 
cool.
 
8:38 AM
@Srivatsan I am aware that displaystyle should not be used in the titles (and I should have noticed it). But when I look at meta question, I have the feeling that for the same reason \sum\limits is not advisable meta.math.stackexchange.com/questions/3135/…
I've noticed \limits in a title in a few other questions, but I left it as I was not sure.
I'm speaking about this question: math.stackexchange.com/questions/52550/…
 
@MartinSleziak Um, has anyone mentioned that \sum\limits is not recommended?
 
@MartinSleziak At least it's not as huge as \displaystyle.
 
@Srivatsan Now, it was never mentioned explicitly.
@JM So I guess I should leave \limits..., right?
 
Also, for some reason letting the summation limits be subscripts and superscripts seems... off.
 
Ok, I'll leave the title as it is.
 
8:41 AM
@MartinSleziak Ok. I had the feeling that \sum\limits is ok, because no one has objected to it. I like that persoanally, but I am ok either way.
 
@MartinSleziak I think so. But, you might want to ask at meta for more opinions if you want.
 
@Srivatsan In the text I would definitely like \sum\limits much more. I wasn't sure about titles (since I wasn't sure how much difference is between this and displaystyle).
 
@MartinSleziak Oh, displaystyle stands out, doesn't it?
 
$\sum\limits_{k=0}^\infty \frac{z^k}{k!}$ versus $\displaystyle \sum_{k=0}^\infty \frac{z^k}{k!}$
 
I don't think it's worth asking at meta, I'm quite satisfied with your advice. You have both copyeditor badge, so your advice is to be taken seriously.
I see.
Thanks.
 
8:46 AM
I actually dislike displaystyle in general, even in main text. Either the math should stay inside the text (in which case displaystyle is unnecessary), or you're better off making it an equation anyway. I guess there could be exceptions to this, but I haven't used displaystyle in any of my posts, as far as I can tell.
Well, copyeditor badge is just about quantity of edits, not quality =)
 
Agreed. :)
 
Almost every time I write $\sum\limits_{k=1}^n$ and $\bigcup\limits_{i=1}^\infty$ when I use it in inline text. Probably a matter of taste, I like it this way.
Quote: Gold Badges are rare. You’ll have to actively work toward these. They’re something of an accomplishment!
But even if it's only quantity, when doing such an amount of edits, there's big probability that if you were doing something incorrectly, someone would have told you about it.
 
@MartinSleziak Agreed. =) I have learned a fair amount of TeX-bits during my edits.
 
@Martin: I suppose, but given the current rate of questions coming in, there's also the likelihood that stuff we screwed up might have been missed...
 
test: $\sum\limits_{i=1}^{n} a_i = \displaystyle \sum_{i=1}^n a_i$
 
8:51 AM
well, \displaystyle only affects what comes afterwards.
or what is your question?
 
No question. I was just trying out what you said in your first sentence.
 
@Srivatsan \displaystyle is what usually appears between $$...$$
 
test: The sequence $\sum\limits_{i=1}^{\infty} f_n(x)$ converges pointwise to $f(x)$, while $\displaystyle \sum_{i=1}^{\infty} g(x)$ converges uniformly to $g(x)$.
[I prefer the first summation over the second.]
 
Heh, I just noticed that you can click "x days visited, y consecutive" in your user profile page and you can see which days count as visited...
 
@robjohn Yes. I thought Martin's claim was that displaystyle and $\sum\limits$ look similar in titles. I disagreed.
 
8:58 AM
@Srivatsan The only similarity is that the limits are displayed over and under.
 
@robjohn =)
 
@tb Wow, I haven't seen that before... maybe it's new?
 
@JM It's been there at least since September-October.
 
@JM I've seen it even under the old profile layout.
 
I mean the thing being clickable. When I hovered on it before, it just looked like text.
 
9:01 AM
@JM Yes. I have been following that for quite some time.
 
Neat, I can see how long my two long breaks took!
 
@robjohn Isn't the approach in this question bogus? The OP is allowing the limits of integration to change with $n$...
@JM The integral $\int_0^{\infty} e^{-x^2}$ is missing a $dx$. Can you edit that? Thanks.
 
@Srivatsan Think of it as $\int_{\mathbb{R}}\;\xi_{[0,\sqrt{n}]}(x)\left(1-\frac{x^2}{n}\right)^n\mathrm{d}x$
 
Is $\xi_{[0, \sqrt{n}]}$ the same as $\chi_{[0, \sqrt{n}]}$, the characteristic function of the interval?
 
@Srivatsan Done, but I was working on it before checking chat. :)
 
9:13 AM
@JM Thanks. [The hope is that you will see it in time before the edit window closes. But nothing is lost even otherwise. =)]
 
@Srivatsan Oops... I meant $\int_{\mathbb{R}}\;\chi_{[0,\sqrt{n}]}(x)\left(1-\frac{x^2}{n}\right)^n\mathrm{d}x$
mixed up chi and xi
 
@robjohn Ok. That's a nice idea... I doubt that the OP has this in mind though (= .
:2709319 First point: what is $a$ and $b$ here? Second: how exactly do we show this?
 
if he can establish monotone convergence, then the convergence is uniform
since all the functions are continuous.
 
Bleh, a and b_n are 0 and square root of n.
 
@robjohn yes, but you need a bit more than Dini's theorem (as claimed in a comment), because Dini needs compactness.
 
9:23 AM
@Matt That would be $a$ and $b_n$... =) robjohn suggested pulling the limits of the integration inside the integral.
 
Better? : )
 
Good morning 8-).
 
Morning Jonas.
 
morning, Jonas
 
@Matt Well, not really. The point is not what name to call the limits of integration. =) It's that that particular theorem the OP quotes uses fixed limits of integration.
Hi Jonas.
 
9:25 AM
@Srivatsan I'd assume he took this question from a text book. "I think I should..." -- maybe not.
 
Hey Jonas and Matt.
 
Yeah, that is true. Since it is not compact, they can't really cite Dini (solely)
 
Hi JM.
 
I love it when things are compact.
 
9:26 AM
@Srivatsan why am I not surprised?
but congrats anyway
 
@Srivatsan Very nice. Do you feel like Indiana Jones?
 
@tb Thanks. You don't sound too pleased. =)
 
@Srivatsan Wow, you beat us to it... :D
 
@JM Yep. I am surprised by that too.
 
@Srivatsan Oh, sorry, that wasn't intended! At all.
 
9:28 AM
Hah. I didn't understand the question. I think now I do.
 
I'm not so much of an archeologist, I think.
 
I'll stay out of this.
 
@robjohn Thanks, rob.
 
@Matt Which Q?
 
Not much of an Indiana Jones fan, really. For a long time, I thought that was a female character. Even now, I find the name awkward. =)
 
9:29 AM
Actually I'm more surprised that it wasn't MaX who got it first...
 
@JonasTeuwen this
 
@JM He went on his editing spree only one or two weeks ago.
 
@Matt Mm, isn't that easy?
 
@JonasTeuwen no Lebesgue theory allowed!
 
9:30 AM
:)
 
@JonasTeuwen Quite the damper, no? ;)
 
Why would you handicap yourself like that? 8-).
 
@JM I think it's because of all the spelling edits and recently the [algebra] edits.
 
@JonasTeuwen : D
 
@Matt Which is exactly what I said, and he is hinting at monotone or dominated convergence.
 
9:32 AM
@JonasTeuwen Yes, like denying the boxer the use of his fists...
3
 
Hah :D.
 
Where does it say "no Lebesgue theory allowed" ?
 
@tb Well, no problem. I didn't take it that way either. =)
 
@Matt Look at the OPs previous questions. And Jonas has this knack of cracking nuts with sledgehammers...
 
(I shouldn't be playing around on SE, I should be doing homework. But this is so much more interesting.)
@tb You accused me of doing that, too. : )
 
9:35 AM
@Matt Well, I'm compensating the tediousness of the government elections with hanging around here...
 
On m.SE it's fun to nuke mosquitoes. On your graded assignments, however... ;)
 
@Matt Yes, I know. But given what you know now, you'd agree, right? (It was applying Riesz's theorem for measures to $c_0$, wasn't it?)
 
Also, I think I learn more on SE than from any homework.
4
 
Wait, why didn't I get this Altruist badge again? No answer, no badge... Did I shell out 300 reps for nothing? =)
 
9:39 AM
@Srivatsan "Altruist" is for actually awarding the bounty to an answer. No answer, no awarding. ;)
 
@Srivatsan I worked diligently on that problem, but I couldn't crack it.
 
@JM Unfair. :)
 
I couldn't figure out how to fit $\sum_k\binom{2k}{k}\binom{2n-2k}{n-k}(-1)^k$ into an inclusion-exclusion argument.
 
@tb I also had this in mind (which sprang from this)
 
@robjohn Ah, thanks. That one is giving all of us a hard time.
 
9:41 AM
This John Klein likes espresso.
I like people that like good drinks.
Plus his comment is funny. +2.
 
@robjohn Um, that perhaps half-baked suggestion of mine. You might as well ignore it and think about Mike's approach, I guess =)
 
@Srivatsan I couldn't think of anything but inclusion-exclusion that would give the alternating form...
 
@robjohn Um ok.
 
@tb I can't seem to find it. : (
@tb Voting is pointless.
 
I like the comment that "Unfortunately FLT is not strong enough to prove $\sqrt{2}$ irrational."
 
9:46 AM
@tb Are you saying all the candidates are utter crap? :)
 
@tb Ha.
It's more recent, I thought it was in an older question.
 
@Srivatsan: your bounty did have the side-effect for me of giving more reputation to others. I devoted too much time to it and it stunted my reputation :-)
 
@JM I don't like most of them. The only mildly interesting thing is that the right wingers plan an attack. Luckily their best candidate was unmasked as a fraud last week, so I don't expect much to happen, to be honest. Now it's mostly rhetorics that last the entire morning.
 
Are you really watching this on telly?
 
@robjohn Funny that you said that. Check my latest comment in that thread.
 
9:54 AM
No, I'm listening on the radio.
 
@Srivatsan you'll see my comment soon :-)
 
@robjohn Saw it =)
The mariner will be rewarded suitably.
 
@Srivatsan Do you get notification (other than the profile page) of comments?
 
@robjohn Not everything. Only those that notified @Srivatsan.
robjohn: If it's not asking too much, can you upvote my comment? I want my generosity to stand out from the actually important comments. =)
 
Does anyone of you have an idea for this? I'd really be interested in seeing that combinatorially.
 
10:03 AM
@tb Nice one. Haven't seen this question till now.
 
10:16 AM
@tb Sounds like a boring thing to do.
I'd like to change "Voting is pointless." into "I throw away all the envelopes unopened." The meaning doesn't change much but it sounds less "unrefined".
 
I'm out for a bit. Later, y'all.
 
See you later, JM!
 
Bye JM
 
Bye JM
 
@Srivatsan Where, other than in your profile, do you get notification of comments (I did mean the ones marked @Srivatsan)?
 
10:24 AM
@robjohn The inbox at the left corner.
And the profile page.
 
Hmm, I could surely use some espresso.
 
Just made myself another cup of coffee : )
 
I demand quality.
There is no good coffee bar in this place.
And my machine is broken. Need to fix it. Will do when I get my salary.
I used to breathe espresso until the machine broke down.
The coffee machine here gives me the willies.
I can't believe people can drink the stuff that comes out of it. It is even harder to believe that they like it.
 
10:43 AM
@JonasTeuwen Dead taste buds. A common phenomenon. : )
But then again, liking whisky also seems like an indication for dead taste buds...
 
Hmm. I don't think so. Whisky has a very broad spectrum of different smells and tastes.
 
I know, I wasn't really serious.
 
Oh :P.
 
I have torture class (Philosophy of Maths) this afternoon. I hope today's student presentation won't be abysmal.
 
@Matt I did that course three years ago 8-). Good luck.
 
10:47 AM
Thanks.
 
loss is always reckoned on cost price and discount is always reckoned on marked price?
 
@Matt This Friday I need to give a lecture in our seminar about the uniqueness of the representing measures in Choquet Theory.
 
@JonasTeuwen Are you prepared? I did my presentation weeks ago.
 
@Matt Not yet.
 
@Srivatsan I need some help in a Q , may I ask the question?
 
10:51 AM
@FreakEnum shoot.
 
@Srivatsan Q: 45 men does a work in 16 days. After working 6 days , 30 more men join them. How many more day they all will take to complete the remaining work?
I came up with this : 45 men's 1 day work = 1/16
45 men's 6 days work = 3/8
remaining work to do = 1-3/8 == 5/8
now what do I do?
 

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