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7:01 AM
Oh, sure.
 
Oh look, it's Martin. My de facto editor. :-)
 
@DylanMoreland I wrote it up just to be sure. =)
 
Wooooooot!
SNOW!
 
@Matt Enjoy!
 
@tb: Look out the window!!
 
7:13 AM
In that case, what the hell are you doing inside? Bundle up, go out and enjoy!
 
@Srivatsan Thanks!
 
We have sunshine outside.
 
But it needs to snow more. About 2 m will finally create peaceful silence.
 
Ok, bbl! : )
 
7:15 AM
@AsafKaragila If it snows in Israel, that would be something. (Just like it was in Venice...)
@Srivatsan Eep, misclick. Thanks!
 
@JM Firstly, in the Golan heights and Jerusalem it snows regularly during winters; secondly, during the 1999-2000 winter it snowed in the town I grew up in (which is not too far from here)
I mean like 60cm of snow, not just a minimal 5cm of snow.
 
@AsafKaragila Interesting; I had thought it was mostly desert-type weather in there...
 
@JM It is.
 
Snow in the desert... can you show me a picture, if you have one? :)
 
I'll see if I can find one. There's one from the winter of 1990-1991 where it snowed about 5cm, but the town was very young and empty. It is an awesome photo.
 
7:23 AM
@Srivatsan Ah, thanks for doing that. I had some stuff written but I'm always somewhat uncomfortable proving facts about permutations.
 
@DylanMoreland You're welcome.
 
Hmpf. Nothing I can find online. It'll have to wait until I'm at my parents and can scan some.
 
@AsafKaragila Sure, take your time. :) (But don't forget me!)
 
I took a group theory course from Bob Griess and we covered a ton of material on permutation groups. I think he thought it would make us better students but I mostly got scared.
 
I am here again for some nagging about tagging.
Could someone have a look at my meta post about tag? If you guys think that it is not worth asking, I can still delete the question.
(At least I think I can delete it.)
 
7:27 AM
I still haven't gotten the hang of permutations. The group theory I was taught when I was a chemistry undergraduate was more of visualizing figures and reckoning from character tables...
 
@JM Sorry pal, it's a slim chance. I'll try though.
 
@MartinSleziak Looks good. I've upvoted the thing I want to happen.
 
@JM Are you a chemistry Ph.D. also?
 
Nope, just a master's degree. I haven't really thought about continuing my education...
(it's expensive and whatnot)
 
@JM Thanks. I basically voted confirmation if it's ok to ask that question. (It seems to me that many questions about tags are recently at meta - I have posted some of them - and often with no result. So I was not sure if it;s worth bringing up.)
 
7:31 AM
@JM Have you made an hypothesis about the continuum of your education, a continuum hypothesis so to speak?
 
Are chemistry PhD students not supported by their departments?
 
Does it happen a lot to you, too: At least two times I wanted to type meta and I typed meat. Some strange typos I do....
 
@DylanMoreland I suck at teaching large classes, which is quite the damper. :D
@AsafKaragila Not yet. Maybe someday.
 
So do many of the people in my year. They run them out there anyway.
I think they try to steer certain people into grading jobs.
 
@JM Just remember that if you cannot prove this hypothesis it may be independent of the axioms ;-)
 
7:33 AM
@DylanMoreland It's a nice subject; I kinda don't want to ruin it for kids...
 
On a related note, @Martin, I think we need an elementary-logic tag as well.
 
@Srivatsan Sounds like a nice fourth option.
 
@Srivatsan I might disagree.
 
Those boolean-algebra calculus questions should go there =)
@AsafKaragila Can you elaborate? I am thinking of it as analogous to elementary-set-theory.
 
@Srivatsan No, I can't elaborate :-P
 
7:35 AM
=)
 
Ok, what else would fit into elementary-logic tag? Some questions about P implies Q is the same as not(Q) implies not(P).
I always forget the English name for this.
 
We already have for that.
 
Contrapositive?
 
Equivalence?
 
What Dylan said.
 
7:37 AM
@AsafKaragila I did not know about that tag. I've added it to the meta post. Thanks!
Yes, I meant contrapositive.
 
No, you meant equivalence! :-)
 
@AsafKaragila That looks fine with me. Except that many old questions that should be here are under , I think.
 
@Srivatsan It is a relatively new tag, so it makes sense.
 
They are discussing teaching some classes in English at our university. (I really don't know who would come to learn mathematics to some second-rate university in Eastern Europe, but the department management hopes to lure some in.)
I am not sure how I would be able to cope with it if I had to teach in English.
 
@AsafKaragila You're right about that. Thanks for pointing it out, I didn't know about the tag.
 
7:40 AM
@MartinSleziak So you've always been teaching in Slovak?
 
@JM Yes.
 
@Srivatsan I often retag when I see fit.
 
I can't read this question. Maybe one of you guys can explain how to interpret?
 
@JM By a vote to close and then to delete?
 
BTW I guess it is the same user: http://math.stackexchange.com/users/21380/marko
and http://math.stackexchange.com/users/20189/marko
 
7:43 AM
He may have meant "numbers" and wrote "nimbers".
 
I think I see what he's asking.
Judging from the question that Martin linked he isn't really into clarifying things with the commenters.
 
@MartinSleziak Do you want to flag the mods for merging the accounts, or should I?
 
@Srivatsan Shouldn't we post a comment to the user first? Suggesting that he should make an account? (Both of them are unregistered.) Perhaps the comment should be posted to the older question, in case this one is deleted.
 
@MartinSleziak No, the user cannot merge accounts. Only mods can.
 
@Srivatsan Yes, but he these both things are unregistered. Shouldn't we nag him first to register and then to merge with these two?
 
7:47 AM
I think he wants to know whether you can take $$\frac{n(n + 1)}2 = \sum_{i = 1}^n i$$ and get a formula for general $\alpha$ that looks like $$\frac{n(n + \alpha)}{2\alpha} = \sum_{i = 1}^n i\alpha$$. And then take $\alpha \to 0$.
 
Of course, you can post a comment urging them to register, but that's different.
 
@Srivatsan Anyway, it seems that you know better what is usually done in such cases, so I'll leave it to you, ok?
 
A nag might be in order, yes.
 
@MartinSleziak You could do that as well.
 
This seems pretty optimistic on his part.
And I don't know why it's tagged "logic".
 
7:49 AM
@DylanMoreland Because it would be "the logical thing to do".
 
@MartinSleziak Actually, why don't you nag the user to register? If they do not respond within sometime, then we can always flag the mods for merger later.
 
@Srivatsan Ok, I'll do it.
 
I don't see where he's getting his new LHS.
You can multiply the first formula through by $\alpha$. But running around replacing $1$s?
 
That was the most bizarre part of the question, I'd say.
 
7:52 AM
@AsafKaragila Yes, but the schmuck's asking for it...
 
Wasn't the nicest thing anyone's ever said.
But there's a history here.
 
So it's bad taste, but not worse than the taste of Victor...
I'll leave it.
 
It's not unlike TG complaining about having to copy a book by hand (if that tale were remotely true).
I'd think photocopying's much cheaper than an Internet connection.
 
Does Victor know about library.nu?
 
7:56 AM
I've managed to formulate a general rule: there's something quite wrong with you if you manage to piss Arturo off...
3
 
Quite.
 
@DylanMoreland I think it looks closer to $$\frac{n(n + \alpha)}{2\alpha} = \sum_{i = 1}^{n/\alpha} i\alpha,$$ i.e., he has $n/\alpha$ summands.
 
I am getting crazy. Again I approved some edit that's way old. :/
 
@Martin Ah!
I should have read more carefully. It was difficult to parse as it was.
So he always ends at $n$. That has the advantage of not being obviously wrong.
 
@Srivatsan I wonder if Benjamin does that to gain the 2 points, and whether or not he'll continue once he hits the reputation to edit on his own.
 
8:08 AM
@AsafKaragila I think he edited one post and then made a follow-up edit on another old thread. =)
But he has made yet another suggestion. I am not touching it this time.
 
So now the idea is to take some $m \geq 1$ and look at jumps of $1/m$. So $$\sum_{i = 0}^{nm} i/m$$?
 
Right. I am going to sleep some more.
 
I don't think that bumping an old question is necessarily a bad thing, at least unless it happens too often. (When I see something in an old/question, which is more just a very minor edit, I do the edit quite often.)
 
Er, no. That's not quite right.
 
@MartinSleziak When he's on the fritz he edits like 43895028753980735 old posts.
Just check his reputation page. It looks like he's trying to get Coffee Editor before hitting the reputation for editing.
 
8:12 AM
When I started with retagging I did many questions and flooded the front page. PM from mods was needed to to get me to some more reasonable rate.
 
So that should be $\frac{nm(nm + 1)}{2m}$, right?
I am always surprised when I can understand something like this. Maybe I'll leave a comment.
 
So I'm gone for the time being. Be seeing you.
 
Goodbye Asaf.
So this formula of his actually works, I think. Incredible.
 
@DylanMoreland I think so.
I got the same result.
Bye Asaf.
 
So he has $\frac{n(n + 1/m)}{2(1/m)}$. And he wants to take $m \to \infty$. That seems dangerous.
 
8:17 AM
For $k=n/\alpha$ I get
$\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2 \alpha = \frac{k\alpha(k\alpha+\alpha)}{2\alpha} = \frac{n(n+\alpha)}{2\alpha}$
 
That definitely blows up, right? When he takes the limit he seems to think that the $(1/m)$ in the denominator goes away.
 
I don't understand that last part of his question, to be honest.
 
Why have I spent any time on this
 
Perhaps there is a language barrier too - his English seems not too good.
@DylanMoreland Well, you can post an answer and gains some reputation... It could be nice feeling to post an answer to question tagged as logic....
 
Haha.
 
8:26 AM
@DylanMoreland Well, if you're not posting what we discussed in chat, I'll do it. If nothing else, it could at least help clarify the question. And as we invested already some time into it, posting an answer seems to be logical - so that our time is not wasted completely.
 
@Matt Woah. Why such a comment? =)
 
@Srivatsan Prompted by jspecter's apparently.
 
Oh, I didn't get that comment either =)
 
@Martin Looks good to me.
I almost edited the title, but I didn't want Joel's comment to become out of place.
 
9:15 AM
@Martin What tags would be suitable for this question I want to know if propositional calculus is the right tag for this...
 
@Srivatsan In my opinion, propositional-calculus seems fine.
 
@MartinSleziak Ok thanks.
 
But perhaps we should wait a little what will come up from the discussion on meta before starting retagging questions with boolean-algebra tag?
 
I am trying to learn by examples. I am not retagging it yet =)
 
9:45 AM
@AsafKaragila It is a bit brutal.
 
@MartinSleziak how do you get the last equation.
 
QED
Seems like people don't like the person rather than the question
 
@robjohn See the answer here: math.stackexchange.com/questions/92190/… Did I make some mistake/typo there?
 
@QED More like he hasn't shown even an iota of effort...
 
QED
9:48 AM
I see
 
@robjohn You noticed that $k\alpha=n$ in the notation used there...
 
@MartinSleziak I see it there, I didn't see it in your comment above. Using $k=n/\alpha$ makes it okay.
Since the easiest way I saw to do this problem was contour-integration, should I add ?
 
@robjohn I think not. The reasoning is similar to the objection for the creation of a Lambert function tag back then...
(I have of course upvoted your answer, but shouldn't you be using \oint in some of those?)
 
@JM The tag exists. I am wondering about adding the tag to that problem.
 
9:55 AM
@MartinSleziak Yes, I remember reading that, and that is why I was thinking of adding the tag.
 
I guess I'll defer to Mariano. :)
 
@JM I specify the integral over the paths $\gamma^+$ and $\gamma^-$. I guess I could use $\oint$ for clarity.
 
QED
Good morning
 
@QED how is your morning going?
 
Hi QED
 
QED
9:59 AM
Great I just remembered I had some nice noodles to eat
How about yours?
 
It's night for robjohn. =)
This is ridiculous! "Lagrange multipliers works" -- almost a nice answer?
I don't even believe this claim.
 
QED
I didn't realize only 10 votes was nice answer
 
You mean that 10 votes is high or low?
 
QED
10 doesn't seem like a lot.
 
@QED My very early morning is going well, thanks.
 
10:04 AM
Someone upvoted tb's comment in that thread just now =)
 
@QED Ya think? I'd be happy to get five for a very technical answer...
 
@QED do you have a lot of answers with more than 10 votes?
 
QED
well I only have one answer with 10 votes - math.stackexchange.com/a/85941/16697
but that's what made me think it's not a huge number, since the answer isn't worth much.
 
That's perhaps why you got 10 votes =) Voting is a bit messed up in this site.
 
@Srivatsan which thread?
 
10:06 AM
 
QED
better answers I've written get less
 
@QED some answers that didn't require much work get a lot of votes, and some technical answers that took immense amounts of work get very few votes.
 
QED
yeah :)
 
@QED Exactly.
By the way, @robjohn, that question seems a little hard [for me], with or without calculus.
Actually with calculus, I am not so sure because I didn't spend too much time.
 
10:12 AM
@Srivatsan the thing I thought to use was calculus of variations, which is essentially what Sunni suggested.
 
@robjohn does it work?
 
@Srivatsan yes.
 
Or are you suggesting an approach? I tried something long back, not sure what I concluded.
 
@Srivatsan: something I missed on first reading of your tangent line answer... "(although this is not a numerical coincidence)" was a nice touch. ;)
 
@Srivatsan I was suggesting an approach, but I've done problems exactly like that one, and the solution comes pretty easily.
 
10:15 AM
@JM Thanks. I was wondering if that remark was vague. Nice to know that you understood and liked it. =)
@JM I am guessing you didn't miss it in your first reading. I revised my answer a bit to improve its flow...
 
@Srivatsan $abcde=1\to\dfrac{\delta a}{a}+\dfrac{\delta b}{b}+\dfrac{\delta c}{c}+\dfrac{\delta d}{d}+\dfrac{\delta e}{e}=0$
 
@robjohn Ok.
 
@Srivatsan Not serious. That's pirate speech, me hearty!
 
The other condition yields $-\frac{\delta a}{a^2} - \frac{\delta b}{b^2} - \frac{\delta c}{c^2} - \frac{\delta d}{d^2} -\frac{\delta e}{e^2}- \frac{33(\delta a+\delta b+\delta c+\delta d+\delta e)}{2(a + b + c + d+e)^2}=0$
 
@robjohn Why are you equating to zero?
 
10:23 AM
@Srivatsan the variation of a constant, which is essentially a derivative, is 0
and the variation is 0 at a critical point.
 
@robjohn Ok, I was missing the second bit. Thanks
Ok, now what do we do?
 
now, we want to have the second equation for all variations that satisfy the first...
 
Aw, right.
How do we enforce something like "for every variation",...
 
QED
I don't see how to continue. What do you do next?
 
It looks like a linear system to me. -- whatever that means =)
Oh, of course. Substitute for $\delta e$ from the first equation in terms of $\delta a, \ldots, \delta d$. (At least in principle)
 
10:29 AM
so linearity requires that we have a constant k so that $\frac{1}{a^2}+\frac{33}{2(a+b+c+d+e)^2}=k/a$ and $\frac{1}{b^2}+\frac{33}{2(a+b+c+d+e)^2}=k/b$ and $\frac{1}{c^2}+\frac{33}{2(a+b+c+d+e)^2}=k/c$ and $\frac{1}{d^2}+\frac{33}{2(a+b+c+d+e)^2}=k/d$ and $\frac{1}{e^2}+\frac{33}{2(a+b+c+d+e)^2}=k/e$
 
@robjohn Oh nice. We get something that looks like Lagrange multipliers, now.
 
@Srivatsan this is how you derive Lagrange multipliers, so it is no surprise :-)
 
@robjohn I know Lagrange multipliers as a blackbox, that's why the surprise =)
 
@Srivatsan That's why I don't usually use Lagrange multilpliers; they are a black box that hides what's really going on.
2
 
So we have $\frac{k}{a} - \frac{1}{a^2} = \frac{k}{b} - \frac{1}{b^2} = \cdots = \frac{k}{e} - \frac{1}{e^2}$. At this stage, in some problems, it will be evident that all of $a$ through $e$ are equal. But not in this one...
 
QED
10:35 AM
What's the meaning of $\delta a$?
 
@Srivatsan there are two solutions for each variable...
 
@robjohn The equation $kx - x^2 = A$ has two roots, say $x_1$ and $x_2$. Then some of $a$ through $e$ are $\frac{1}{x_1}$ and the remaining are $\frac{1}{x_2}$. Here $A$ is some undetermined constant. [I am not sure if you want to take this route though]
 
QED
(Surely a,b,c,d,e are functions of a point in 5D space, rather than functions of paths?)
 
@QED Not sure what you mean =)
 
QED
I'm just lost trying to follow this
 
10:37 AM
@QED $\delta a$ represents a small change in $a$. For any change of the variables, we have $a=a(t)$, $b=b(t)$, ... and $\delta a = a'$ etc
 
QED
so it's just the normal derivative?
 
@QED it's a rate of change, so yes, a derivative.
but to be taken over any of a number of paths.
that's why I say, "for any variations".
You want the change in your value to be 0 for any of a family of variations of the parameters.
@Srivatsan However, that they each equal $\frac{33}{(a+b+c+d+e)^2}$ usually means they must be equal. I will have to think a bit.
 
11:03 AM
@JM, I can't follow this comment at all. Am I alone in this or you can't follow also? =)
 
11:30 AM
@Srivatsan I've replied. If that doesn't get to him, nothing will.
 
@JM I think he's shifting the goal post of what's intuitive...
I guess I'll be giving up on that question soon =)
 
@Srivatsan I agree.
 
But thanks for the comments =)
On a different note, do you think this comment will have any effect? =)
 
I hope so.
 
@JM looks narqy to me
Hi all
 
11:34 AM
@tb I'd cast my vote if the next comment is screwy enough.
and hi. :)
 
@JM I too. It's kind of irritating to find that answer with 9 votes -- I am one of the culprits :(.
hi tb
 
Bah, the comment didn't work. His "intuitive" isn't the same as mine.
 
Hello tb : )
 
Hi Matt
So, got your moon boots ready?
 
hi Matt
 
11:41 AM
@tb : D What makes you think I have any?
Hi Srivatsan.
Minus one puppy. : )
 
@Matt It was more, like, a figure of speech :)
What do you mean? Did you give the puppy away?
 
Yes.
 
I'm back!
And I feel like having a burger.
 
@tb No, we made dog stew.
 
@Matt I try to avoid eating animals that are too far up the food chain...
 
11:48 AM
I explained the class about minimal vs. minimum with the food chain. I told them that lettuce is a minimal element there because it eats no one, but it's not the minimum because I don't eat lettuce.
 
@AsafKaragila ...in the usual food-chain order, right?
 
@Srivatsan Of course.
 
@tb Are you a vegetarian? I'm asking because the lasagne contains things that are higher up the food chain than lettuce.
 
@Matt What would that be?
 
Well, I somewhat see the point of avoiding carnivores in your diet...
 
11:57 AM
@Srivatsan Beef, pork and anchovies.
 
Yuck! I hate fish.
 
@Matt Not at all. What J.M. said.
 
@Matt Really!? They never told me this... =)
 
@Srivatsan It's not proper lasagne, it's a Jamie Oliver's creation.
 
@Srivatsan Matt said the lasagne, not all lasagne.
 
11:59 AM
@Matt oh, cool.
[Not that it matters that much.]
 
Can someone wire me some money so I could go out for burgers?
 
QED
er
binary trees with nodes of degree other 2?
 
QED
1
Q: Prove $n_0=(k-1)n_k+(k-2)n_{k-1}+\dots+3n_4+2n_3+n_2+1$

GigiliI want to prove $$n_0=(k-1)n_k+(k-2)n_{k-1}+\dots+3n_4+2n_3+n_2+1$$ for T as a binary tree where $n_i$ is nodes of degree $i$. I tried to prove it using the handshake lemma but came up with nothing useful.

I don't get this
looks like "binary" should removed, from the comments
 
Well, Brian is about to answer, so just wait a few minutes.
 
12:10 PM
hi
 
QED
edit it to say womyn instead
 
@Srivatsan I'm wondering whether giving $-1$ would be unintuitive...
 
@tb hilarious =)
 
Hi, Rajesh
 
12:26 PM
I have downloaded a book trigonometric series only to find later that it is in Russian !
4
hi @tb
 
Who are the authors?
 
Zygmund
and Bari
Zygmnd is american i guess
I am being bugged by this theorem in Fourier series...hope someone could help me sort it out
 
There is an English book by Zygmund, published by Dover.
 
i have a copy which is old but i found a latest edition on google books
so thought of searching for it
@JM discussing pirated stuff : is it banned here ?
 
@RajeshD Let's not go there. There is a new American law about to come up that might put this site at risk.
 
12:34 PM
oh my ...ok
@JM do you have a hard copy of that book with you now ?
 
No, sorry. I'm far away from my library.
 
ok
i need some help in sorting out a theorem in it
hope someone around here could be of help
 
Unfortunately, I need to leave now. Sorry. I hope somebody can take over from me.
Later.
 
k bye
 
QED
2
Q: proving you *can't* make 2011 out of 1,2,3,4: nice twist on the usual

Kevin BuzzardAn undergraduate was telling me about a puzzle he'd found: the idea was to make 2011 out of the numbers 1,2,3,4,$\ldots$,$n$ with the following rules/constraints: the numbers must stay in order, and you can only use $+$, $-$, $*$, $/$, ^ and $!$. In words, "plus minus times divide, exponentiation...

that's a tough one
if you consider the values of the set of expressions with 0 factorials, 1, 2, ... it's always going to include new numbers
So one will need to get up to a point where they can prove the cancellation is separated by a big enough gap
 
12:47 PM
Nice. Port with blue stilton and a fig.
 
QED
that sounds great
 
It is great.
 
see you all.
 
QED
bye
 
Bye Srivatsan.
 
QED
1:14 PM
So do the variational derivatives tend to prove these rational function inequalities automatically?
 
@robjohn I use Lagrange multipliers in infinite dimensional spaces :-).
(with inequality constraints)
 
1:37 PM
@QED no, they take some work.
I killed QED!
I replied to QED and I saw their avatar pass away :-(
 
Pheew, one more of those answers that took waaay too much time...
@robjohn Looks like QED's resurrected
 
@tb Which answer?
 
@robjohn this one
I usually don't like it when people assert that there's a mistake when obviously there isn't.
 
@tb OMG that is ginormous!
I do not understand what you are saying, but it looks impressive :-)
 
1:54 PM
@robjohn Well, I guess the reason people don't write that stuff up in detail is that it is not really difficult, but writing it out makes things explode in size.
 
It took me a while to do a contour integration, but I had not seen the problem when it came out a month ago. I think the bot must have regurgitated it since there was no reason it should have been at the top of the list this afternoon.
@tb Isn't there a way in $\LaTeX$ to make diagrams like that?
 
I admire Krause's and Keller's ability to compress everything to the bare necessities. When you fill in the details and go back to their texts, you see that they say exactly what you need to do...
@robjohn I use xypic for that (that's how I made the pics on my laptop). Unfortunately xypic and tikz aren't implemented in MathJaX (probably because the code of xypic is a hideous monstrosity). See the links I gave in the first comment here for more on that
 

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