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00:00 - 21:0021:00 - 00:00

00:00
@DylanMoreland I don't know, I never bothered, but those who did, manage to do a lot of good stuff with it. On the other hand, I think with a bit of practice XY can draw pretty nice diagrams, too.
QED
QED
L-system for the fibonacci sequence
plus if you view A as "left 1 step" and B as "up 1 step" I think it is a path that always stays below a line of gradient phi
@QED Well, the trend is more or less exponential in the long run, so I don't see how it'd stay below some line...
I'm half-tempted to tag this ... :D
heh :)
@JM what do you mean by "exponential in the long run"?
I would have said this thing behaves more or less like the lower convergents of the continued fractions expansion of $\phi$
00:15
@tb for large enough index, Fibonacci numbers are nearly indistinguishable from $\phi^n/\sqrt 5$, so I'd think a plot of these with increasing index wouldn't stay below some line.
@JM What are we plotting against what? If it's the number of A's against the number of B's, then it probably converges to $\phi$, no?
@Srivatsan On the other hand, I was making assumptions on the other coordinate QED was using, so I could be wrong. :)
@JM Actually, I am not also not clear. I was guessing it.
I thought of the $(x,y)$-plane and the line with slope $\phi$. Basically you're going $f_{2n}$ to the right, $f_{2n+1}$ up
Ah, that works.
00:22
@Srivatsan Hows it going?
@Sivaraman Are you back?
Not yet - I'm in Istanbul - leaving tomorrow - should be back in ~36 hours :)
Wow, Turkey.
:) Yes
@Siva Ok. I'll see you then. If I am in this city, i.e.
00:24
Oh, Istanbul, I envy you!
@Srivatsan Where are you planning to be?
@tb It is a great city :)
@Sivaraman Not planned. Perhaps Allentown, PA. My cousin's.
@JM Yes, he's my hero. =)
@Srivatsan If you're around and up for it let me know and we can get food sometime?
00:28
@Sivaraman Sure. :)
@tb How come Pittsburgh does not trigger any envy? =)
@Srivatsan "I don't know if it was intended that way, but the tone of the response to Srivatsan is very hostile. –" I'm very tempted to say youtube.com/watch?v=AKtwlHV1-O8
@Srivatsan I can answer that :)
@Sivaraman I hope not. =/
Ok, I'm feeling hungry. I should leave...
@Srivatsan You need to switch your screen name to Shane/Salty
Ciao :)
Salty would work.
Salty out.
Ciao Salty :)
00:34
Since those three guys have been brought up, I'll take the opportunity to link to this...
sweet :) although I didn't get the reference.
No references, it was just something those three came up with. :) Nowadays people use it for vocal exercises for choirs.
You said "those three guys have been brought up". That reference I didn't get because I overlooked Sivaraman's link...
Oh, that. I just like the Stooges. :)
I can't tell, I don't know much of them.
00:50
@Ilya It looks as if it is already opened. Sorry, I was gone today, so I didn't get to it earlier. I tried to forestall the closure, but I was unsuccessful, it seems.
@Sivaraman Salty?
@robjohn well, we triggered closure and reopening because it gathered a fourth vote for closure.
welcome back :)
@tb Thanks. It's been a busy day.
@robjohn I'd be surprised if it weren't. :)
Yeah, you mentioned going to the vet, so I'm curious... How's the kitten's hip?
I see that none of my old answers were upvoted today. Not that I expected any, but sometimes it happens.
@tb It is pretty much healed. We can let her out for an hour at a time.
@tb however, the vet didn't want her let out all the time yet.
@tb We're supposed to take her back in 6 weeks.
00:55
@robjohn That's good news in any case. Considering that it took so long already it's probably better to be on the safe side.
@tb The x-rays looked pretty good and the kitten is not showing any signs of any problems.
@robjohn well, yesterday I had negative score and today one upvote (and a non-answer that I deleted because I was too lazy to fix it and now someone else wrote it up).
@tb I hate it when that happens. :-(
I simply misread. A Cayley transform would have done the trick but I always mess it up... See here
(the negative score was due to my voting down, so I'm not feeling very bad)
@tb Yes, I see that. Not that you always mess it up, but that a Cayley transform would be applicable.
01:18
@robjohn there's an even simpler fix: just put $f(\bar{z}) = 1/\overline{f(z)}$ for $z$ in the reflected region $S$
(I should have seen that immediately)
01:59
@tb Sorry, I had to go afk to deal with someone at the door. Now I need to walk the dog. No rest for the wicked...
02:12
@robjohn Er, that would be me... =)
Be back in a few. Later.
user20683
user20683
Merry(Happy) Christmas

Let all your presents be provably awesome and let set C = {Joy, Cheer, Good, Food, Peace}
@WorldEngineer Thanks, Merry Christmas to you too!
03:30
@tb, About the convex function question, wikipedia seems to think that composition of convex functions need not be convex. The outer function should be monotone increasing...
@Srivatsan whoops, you're right. But the inner linear is good enough, too :)
Um, interesting.
For the compositions you can take something silly like $(x^2- 1)^2$, of course
I don't get this. This is the outer function, right?
Inner function $x^2 -1$ is convex, outer function $y^2$, too, while $(x^2-1)^2$ isn't
03:38
Oh, I like this counterexample actually.
It's probably one of the simplest possible examples: you need both functions to be non-linear.
@Srivatsan And what you call Jensen is just the property $f((1-t)x + ty) \leq (1-t) f(x) + t f(y)$, right?
(that's what I was trying to point out in my comment)
Yes, that's Jensen. (No?)
Okay, for me it's the inequality $\phi (E[f]) \leq E[\phi \circ f]$
What's $\phi$?
a convex function
03:48
Oh, right. I thought $f$ is the convex function, but that doesn't make sense. What is $f$?
Oh right. $f$ represents the probability distribution, is it?
A measurable function (a random variable). Should I have written $X$ instead. I'm referring to this
(where your thing is the finite case and my thing is the measure-theoretic version)
Yes, I get it now. Yes, my Jensen is the baby version of yours.
@Srivatsan well, "mine" can't be proven without "yours", I guess. But I use that one as the definition of convexity.
By the way: do you know how to prove that among all $n$-gons inscribed in the unit circle the regular $n$-gon has maximal area?
It will suffice to show this, no? If $\stackrel{\frown}{AC}$ is an arc and $B$ is a point in between, then the area of the triangle $ABC$ is maximised when $B$ is the midpoint of the arc.
I'm wondering if using this inductively finitely many times will show that the given optimal n-gon is regular.
@Srivatsan I'm not sure I understand what you say
04:03
One second.
(but this wasn't a random question I just happened to throw in completely out of context...)
Lemma: Suppose $\stackrel{\frown}{AC}$ is an arc, and $B$ is a point in the interior of the arc. Then the area of the triangle $ABC$ is maximised when $B$ is the midpoint of the $\stackrel{\frown}{AC}$.
Does this statement make sense?
Yes. And it's obviously true because then the height is maximal.
Now suppose that the given $n$-gon of largest area (assume this exists...) has some two sides unequal.
Okay
Hm. I don't see where you're heading.
04:12
@tb Sorry, got disconnected.
Oh, no problem :)
Suppose that there is an $n$-gon of largest area. Assume it has two adjacent sides $AB$ and $BC$ (say) unequal. Focus on the arc $ABC$ and apply the lemma: you can "correct" the $B$ to $B'$ such that $AB'C$ is of strictly larger area than $ABC$.
The rest of the polygon is the same. So we got an even larger $n$-gon, contradiction.
Oh, nice!
[And the largest $n$-gon should mostly exist by compactness.]
Yes, but I don't quite see how to prove this...
04:18
Neither do I =) . But let's try.
One second.
A polygon can be identified with a list of $n$-points in the circle, sorted (let's say). So the set of polygons is a closed subset of $(S^1)^n$.
Yes. // no, the admissible points are distinct, so it's not closed.
@tb Relax that condition =)
Let's allow degenerate $n$-gons which are really $k$-gons for some $k \leqslant n$.
Okay
It's fine, proof complete by continuity.
Perhaps even sorting is unnecessary.
@tb Damn, I was totally going to say "Ha, you missed continuity" :(
=)
Here's what I had in mind:
Certainly a polygon that's a candidate must contain the center of the circle.
Now draw lines from the center to the vertices, get $n$ angles $\alpha_1,\ldots, \alpha_n$ (at the center).
We have $2 \pi = \sum_{i=1}^n \alpha_i$
04:27
Breaking news: New question posted by V.
@tb Sorry, go on.
The area of the polygon is $\frac{1}{2} \sum_{i=1}^n \sin{\alpha_i}$
Maximise $\sum_{i=1}^n \sin \alpha_i$ subject to the budget $\sum_{i=1}^n \alpha_i = 2 \pi$.
Yes, and our assumption on the center being in the polygon tells us that each $\alpha_i$ is actually in $[0,\pi]$, where $\sin$ is concave.
Yes. That solves it also, right?
So, $\frac{1}{2} \sum_{i=1}^n \sin{\alpha_i} \leq \frac{n}{2} \sin{\left(\frac{1}{n} \sum_{i=1}^n\alpha_i\right)} = \frac{n}{2} \sin{\frac{2\pi}{n}}$
04:31
Nice...
@tb Is there a catch anywhere? The proof is ok, right?
No catch. It's my favorite application of "Jensen"
(that's why I brought it up)
Ah, um. Yes. Nice... =)
Thanks for this.
But your proof is even better!
Thanks for this.
QED
QED
0
Q: Sum of two cubes

VassiliDoes anyone know how to get all solutions of the equation $a^3=b^3+c^3$ when $b,c$ quadratic irrationals or $b$ rational and $c$ quadratic irrational?

can some mod delete this so I can re-ask it?
(I want to know the answer)
Someone is swapping all occurrences of $a$ and $x_0$ for some reason =)
Actually not. There's some general confusion in the answer.
04:41
@Srivatsan yeah, the assumptions in the question are continuity at $x_0$ and desired is the continuity at any other point.
I have no idea what the suggested edit is doing. I think the answer is fine, but I want to understand the suggestion. That seems very hard in this case...
too late :) QY just approved it.
That one is so hilarious. Why would anyone want to switch those two arbitrary symbols?
In order to match the answer with the assumptions in the question. $x_0$ was given as the point of continuity of $f$.
The question mentions neither $a$ nor $x_0$.
Oh, it mentions $x_0$. I missed it, my bad. Now it makes sense.
04:47
well, actually the edit I don't really understand is this one
@tb Oh that explains it. jspector probably did not see Paul's edit.
@Srivatsan yes, probably
I'm tempted to vote to close this one as too localized.
Hi, 2,343,432,205! Funky name...
OK, see you.
05:00
See you!
Hello
where $x$ in a plane coordinates of a given
image
Is this sentence right?
I mean 'a plane coordinates'
Hi: I'd write "where $(x,y)$ are planar coordinates of a given image"
where $x\in\Omega\in R^{2}$ in a plane coordinates of a given image.
t.b.?
How could I change it?
Thank your very much.
$t.b.?
@t.b.?
I'm not quite sure what you're trying to say: is that image a mathematical object or is it an actual image? (in any case write $x \in \Omega \subseteq R^2$.
Actual image?
Yes, it is a actual image? My major is computer vison
vision
05:30
I think you're trying to squeeze a bit too much information in one sentence. You should say that you work with planar coordinates in $\mathbb{R}^2$. Then say what $\Omega$ is. Then you can simply say that "where $x \in \Omega$ is the location of the given image"
or "where $x \in \Omega$ describes [or: determines] the location of the given image"
Yes, I am.
 
3 hours later…
08:33
Good morning everyone!
How do you do?
@Martin: It seems that we drove that guy away... :-P
You mean Matt?
I've seen you mentioned I. Farah.
No, I meant you. The guy that didn't want to accept any answers.
What guy? Just making some coffee.
@MartinSleziak Yes, I met him yesterday. He's a very nice person.
@Matt I'm good, thanks.
08:37
I have too books by him. One is "Analytic Quotients" another one is about forcing (joint with Todorcevic). I hope I will get to reading the books some day.
I believe you mentioned topic approximately like in his notes here: math.cmu.edu/~eschimme/Appalachian/Farah.html
@AsafKaragila Was that question deleted? I cannot find my comments. (Was his name SVV or something like that?)
@MartinSleziak This is the question. The user was deleted, then the question undeleted but the comments were removed.
Weird. But maybe the pressure on accepting answers was too big recently (e.g. Gerry's answer; but that was a different user).
@MartinSleziak I disagree with that. Especially, in this case, in light of how he replied to me, and the following correspondence with you.
@AsafKaragila I only wrote that accepting can be useful, so that the question does not pop out on the main site repeatedly. He admitted this might be a reasonable point, but that this seemed different to him from what preceding comments were saying.
He didn't want to follow the norms, when you gave a very good reason for accepting he agreed with that - but since it is not what the other users said he will not accept answers... that's just standing on the principle of things without a good reason.
08:46
Well, there are plenty users with low acceptance rate. I am just saying I am not too happy that he went away from the site (that the user was deleted).
I know there are many, often they do not know about accepting answers and rectify that when being told. However he was a bad case of "Never do what you're told." and preferred to leave rather than trying to understand.
Do we want separate tags for and ?
They're different things...
Yes but they're related.
ok, thanks Zhen. I wanted to here other peoples opinion.
08:55
"almost groups" are different too...
Oh shush.
I'm disappointed with the new version of Liferea.
You guys use RSS readers?
I use Google Reader.
I have no intention of bringing Google into my RSS.
09:07
@JM : D @Srivatsan You guys are thinking way too much : D They were not being silly. I also wear their trousers sometimes because they're baggy and that's comfy : ) And yes, they were offering new clothes.
@Matt I started reading back on the replies of the replies of this message, and I got to the strange question "Do you think you can fall in love with someone that you met online?"
From that to this is quite a digression, and to think I wasn't even involved in the conversation!
@AsafKaragila Poor you.
@Matt Yes! I demand to be a part of every digression within this chat. Especially from one general nonsense to another.
3
That wasn't nonsense. That question was addressed to Ilya when I asked it and then I decided to remove the ping.
@Matt Oh.
09:16
And I think where it now says "Hell I deleted the right post" could've been what I was referring to.
Dammit. I have a homework assignment due by Thursday, but whenever I start working on it I find myself working on my thesis instead.
What subject?
Commutative algebra.
10:06
Time to go to the Japanese bakery to have some yummy breakfast, bbl : )
@Matt Enjoy!
10:47
@Matt Japanese bakery???]
11:45
In [this](http://math.stackexchange.com/questions/82240/proof-of-the-lebesgue-number-lemma) one I think you can delete the 3 first sentences in the second paragraph.
For the inclusion argument I think I'd choose $\delta := \frac{\min r_x}{2}$ and then because $d(x,y) < 2 \delta$ you get $B(y, \delta) \subset B(x, r_x) \in \mathcal{A}$. What do you think?
(@Srivatsan)
12:10
@Matt how are you doing?
12:29
@Ilya Fine thanks, and yourself?
12:53
merry mathmass!
Happy halloween =)
13:23
Oh right Dec 25 = Oct 31
@Matt myself is good
Good. : ) Did you find fast food for dinner last night?
@Matt no, so we went to pizzeria. how was your lasagne?
It was good. : )
13:47
So... who wants to solve some commutative algebra with me?
14:44
I would actually but I wouldn't be of much help.
@AsafKaragila :-)
@robjohn We'll see. :-)
@AsafKaragila remind me which second course do you have? algebraic topology?
15:05
@Ilya Yes.
 
1 hour later…
16:20
Well, I can't stay long since I'm the designated cook, but: Merry Christmas!
(BTW, I'm making lasagne.)
@JM Isn't it almost midnight...?
17:21
@JM : ) Nice. Have fun with the cooking!
17:35
I thought xmas was tomorrow.
17:49
Merry Christmas, all of you. Have a beautiful day.
18:10
I have proved that dx/dy is indeed a ratio! If it wasn't why would you use \frac{dx}{dy} in LaTeX to produce it?? BAM!! Proof by LaTeX!
5
@Matt Far enough east (UTC+6) it is Christmas now.
QED
QED
Can't you just ban the guy?
instead of unfairly downvoting his questions
QED
QED
-1
Q: Sum of two cubes

VassiliDoes anyone know how to get all solutions of the equation $a^3=b^3+c^3$ when $b,c$ quadratic irrationals or $b$ rational and $c$ quadratic irrational?

Merry Christmas @JM!
@QED you brought this up yesterday. Do you have an answer?
QED
QED
18:23
I've read research papers before that solve the problem for Q(\sqrt{n}) for some n but not all n and I don't remember the authors or anything
@robjohn Oh! True! Merry Christmas, @JM!
QED
QED
I'd like an answer
Actually, @robjohn, what are your ties to Japan? Isn't it slightly unusual for a gaijin to like genmai cha?
@Matt ...said that gaijin!
18:36
@QED I'd like a back rub.
QED
QED
In return for what?
I'll be back later to negotiate
@Matt You're Swiss, that's not a clan of Ninjas as far as I know.
@QED An answer.
Is a question about heap on topic? Hello.
@AsafKaragila : )
Hip bones?
18:45
@Gigili Sure, ask away.
Uhum, thank you @Matt. Which tag should I use? (tree)
The leg bone connected to the hip bone, the hip bone connected to the pelvic bone, the pelvic bone connected to the spine bone, the spine bone connected to the backbone, the backbone connected to the neck bone, the neck bone connected to the wish bone, the wish bone connected to the fairy bone, the fairy bone connected to the gay mafia, the gay mafia connected to the film industry, the film industry connected to the Oscar, the Oscar is the first name of Zariski...
@Gigili Oh, I thought you meant here in chat. On SE I wouldn't know which tag. Try with heap, the worst that can happen is that you get migrated to SO.
@AsafKaragila You on drugs again?
@Matt The Prisoner is on drugs as a series, I didn't have to do anything :-)
QED
QED
The Prisoner was damn good
18:50
This is what my university homepage has.
@QED Have you read Shattered Visage?
QED
QED
no
I'm not sure if I want to or not. I mean, it was so awesome how the series ended that I don't want to spoil it. On the other hand, I have to know...
QED
QED
Oh cool
It's like a sequel type thing
It's a graphic novel.
Also, there is a 40 minutes long awesome metal opera called The Girl Who Was... Death, from which I originally learned about the series.
The band's name is Devil Doll.
19:09
@Matt "Suppose a max heap with N different elements which is implemented by array (the biggest element is the first one in array), what's the index of the fourth biggest element?" If you have time to think about it , please ping me so I'll read your answer later. Thanks a lot.
19:29
@Gigili There is more than one valid index for the 4th largest element. Take for example N=4, then you get 3 valid max-heaps and if you look at the index where the 4th largest element is stored you'll find that it could be array position 2 or 3. For bigger N you get more possibilities. Are there any further constraints on that heap?
19:51
Wat das "Auswahlaxiom" mean?
How's Christmas Eve there, Asaf?
Thanks. :)
@AsafKaragila Axiom of choice.
@Srivatsan Just another Saturday night.
@Matt I see. Thanks!
I recently came to an understanding: The axiom of choice depends on your choice of axioms.
20:04
: )
Asaf: Oh, that's good. It's worse when the entire world around you celebrates what's just another saturday night for you. :/
Merry Christmas, @Matt.
@Srivatsan I'll ask you around Rosh Hashana, or passover how you're doing :-)
@Srivatsan Thanks, merry Christmas to you too! Although technically it's not Christmas yet : )
@AsafKaragila I might actually know that some festival is on. My neighborhood is a Jewish one.
@Srivatsan Fine. I'll ask you in the Israeli Independence Day :-)
20:10
:)
20:42
@Matt Could you please explain it to me?
I don't get it, even for N=4 *=* .. But your answer is correct according to my book, it says there are possibilities from 2 to 15 index) where the 4th biggest element is stored.
@Gigili Ok, assume you have the keys 1, 2, 3, 4. Now write down all valid heaps.
See for example here to look up the conditions that a heap has to fulfill.
Barney Heap?
For example, every level of it has to be full except for the bottom level.
Put the maximal element into the root (index 0 in your array)
Then make sure every child is smaller than every parent.
And if the bottom level is not full be aware of the fact that you have to fill from left to right.
@AsafKaragila A sort of data structure. Although I've never implemented one in practice.
20:59
@Matt I got 98 in the Data Structures course. :-)
@AsafKaragila I got a fail. I ignored it. Dunno why I did that because it counted double.
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