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20:03
@r9m That proof blows up minds. I'd like to be the problem on the cover of my book.
@r9m woah, nice proof indeed
@Hippa Cole, Dunn, & Filaseta (2015?).
That's the paper.
What paper ? (so many thing happened i can't remember) @BalarkaSen
@Hippalectryon They prove the currently open problem on deciding whether it is possible to have a polynomial $f(x)$ with nonnegative coefficients bounded above by a given $n$ with $f(2)$ prime.
20:05
I found an example with $n = 10$, remember?
I remember now
r9m
r9m
@Chris'ssis :D .. awesome news .. so you have started writing the Book after all ?! :D
@r9m Not really, but that will be one of the problems in my book. :-)
@r9m You mean Erdos' "The Book"?
OK, OK, I am going @TedShifrin
20:08
You keep saying that xD @BalarkaSen
r9m
r9m
@Chris'ssis :D
:O a wild wolf @TedShifrin
Just reported upcoming paper above, @TedShifrin
r9m
r9m
@G.T.R :-)
20:08
You don't know, @Hippa. Glad to see you haven't burnt to a crisp yet.
Out, @Balarka.
@TedShifrin My left hand hurts (the left side is where the hot part is and where the hooooot air is blown out) though
Seriously ... Get it fixed or you'll have no computer. Point.
what kind of computer is it?
A melted one.
hi everybody btw
:-)
20:11
@TedShifrin Say it again then i'll be able to draw a line :D
That point was French.
And so ?
:P
@r9m did you manage to finish this one? $$\int_0^1 (\log(1-x) \log(1+x))^2 \ dx$$ It's awesome too!
Je n'ai point d'idée.
r9m
r9m
@Chris'ssis nope :(
20:13
@r9m Maybe you don't like it that much ... (that's why)
@Chris'ssis Why are lots of your integrals 0->1 ?
@Hippalectryon not really "lots".
Ah my bad
r9m
r9m
@Chris'ssis I must admit I don't have a deep attraction towards evaluation of integrals .. but never I have said I don't like it :P .. its perhaps too strong for me imo :|
@r9m That's rehashed but still: Prove there is no function $f:\mathbb R \to \mathbb R$ such that $\forallx, f(f(x))=x^2-1996$
20:15
@Ted There's no constructive proof of Riemann mapping, is there?
@G.T.R Doesn't it work with $2014$ ?
Sure
@Hippalectryon $$\int_{ -\infty}^{ \infty} \frac{ x^2 e^x \operatorname{Li_2}(e^{-x})}{(e^x-1)^2} \ dx$$ that i created it last days. All it depends on the direction my research pushes me to.
@G.T.R Then put $2014$, it's better :D
r9m
r9m
@G.T.R this is a very old IMO problem .. as far as I can remember (If memory serves me right)
20:18
@r9m your memory fails then :P
@Chris'ssis I'm saving that one too :D
@r9m Let me guess with my magical powers.... it's from.... 1996 ? :P
Probably there is, @Mike, using PDE/Green's functions.
@Hippalectryon The one I just posted above? :-) It's AWESOME . :-)
@Chris'ssis yeah - i've got like 20 screenshots of your integrals
@Ted When I say constructive, it's not a very mathematical term. I mean something that actually constructs a map... and that if I work far back enough, I can actually write down a map if I have a region.
20:19
I'm storing them so that when I get some time/knowledge i try them
@Hippalectryon :D
For now i can't do them yet
I imagine that using PDE or green's functions would have trouble starting from writing down the green's function.
That's why your computer is burning, @Hippa.
4
@Hippalectryon That's nice to read! :-)
@TedShifrin lol :-)))))
20:20
@TedShifrin What the link -_____________-
Yeah, probably, @Mike. I suspect the normal families proof could be implemented on a computer ...
I might look into that.
I've been trying to do some stupid computations for a while now :P and I have all these nice theorems at my disposal, saying that "This computation can theoretically be done"... while giving no tools for doing the computation.
@Hippalectryon have you done a search on Youtube to find methods to fix your computer?
@skullpatrol I didn't find any, except cleaning the fan, which I have done already
That's a good start :-)
20:24
I still have 99°C temps spikes
Average 92
I think @GTR's video of the Russian roaches is probably right.
You could post the question on Computers.SE
hello everyone
It's a laptop, they couldn't even get in @TedShifrin
20:25
@Ted But at the same time, if I could implement the normal families proof, I should in principle be able to just use Perron's principle to solve the Dirichlet problem on my domain... which is all I want to do anyway. I don't know how to write down all the elements of the normal family.
Suarez the vampire vine.co/v/MtJwm0p3Jwh
And they'd get fried instantly
Hi @Lucio
OMG @WillJagy hi!
@G.T.R Yes I was that too ! It's so....... If not for the video, i would hardly have believed it
20:26
Can you make references to MSE posts on chat or is it frowned upon?
I'm tempted to reference an MSE question to answer that, @Lucio
Is @Will here? berkeley rules :)
(I don't think anybody cares as long as you're not spamming.)
:) Okay...
@TedShifrin hi. Got a very satisfying answer on a field theory item on MO, original idea was Jyrki suggesting a blog post. Maybe
20:27
loooong time no see :D
You're blogging?
If someone has a chance could they check the question under the post 'Non-divergence form of a 2nd order PDE', the question is by Alex...
BTW, I'll be in your neck of the woods end of July @Will.
Any assistance on that question would be appreciated...
Better to link to it, @Lucio. Don't know how many PDE experts we have
20:30
@skullpatrol I don't think they could help me
@TedShifrin You'll be happy to know I've realized I'm an idiot
2
@Ted, good. I'll be here. Sometimes people have time to socialize when they visit here, sometimes not. i am in the telephone book, should you find one, also ams.org/cmlgives my home number
@Ted It doesn't require a PDE expert...should I coy and past the url?
@Mike, you do that 10 times a day.
@Will: What is a phone book?
I remember being suspended for doing that once...
20:32
@TedShifrin This time it was because, following along a certain proof that uses the Riemann mapping theorem, it's constructive up to having the Riemann map... but for the regions I'm considering I have a Riemann map fairly easily. (The simplest ones I'm doing right now are just the (infinitely) slit plane!!!)
we use the [subject](link) format a lot here, @Lucio.
@Ted how exactly do I do that?
@Ted, about a year ao I was talking on a bad cellphone connection with a young woman who lived in my building with her husband a year or so before. i said I'm in the phone book and she got indignant, she said I don't have a phone book, the very thought offended her.
Can I just paste the url?
LOL, I don't get indignant ;) My visit is entirely social, visiting lots of old friends. Not totally sure where I'll be where ...
20:35
@Hippalectryon it doesn't hurt to try :-)
@skullpatrol It burns :P
@Hippalectryon Does stackexchange even have a site for computer repair?
@skullpatrol I don't think so xD
@Hippalectryon have you looked?
@skullpatrol I've looked at so many websites, I doubt SE could bring any other thing
20:38
ok
@TedShifrin, I'm trying to understand the proof for Proposition 2.1 here: people.reed.edu/~jerry/361/lectures/lec07.pdf
What do the mean towards the end when they say "whose additive counterpart..."?
@Ted, the one miss I sort of regret was a Swiss man with the last name Jaggi I met at the January 2013 combined meeting in San Diego. Cannot be sure, and my father had other ideas, but I suspect my last name is a corruption of Jaggi. Anyway, he was here for a few months but I did not look him up. Having studied German, the correct pronunciation of Jaggi is rather surprising.
@KajHansen How do you create a link like that?
@KajHansen right above they gave an isomorphism between a multiplicative group and an additive group. the equation $x^d=1$ in a multiplicative group is the same as $dx=0$ in an additive group.
20:41
They mean exponential in an additive group corresponds to multiple.
@Lucio: That's what I told you above.
@Hippalectryon have you tried here?
well, bye
@LucioD, what do you mean? I just copied/pasted a URL.
@skullpatrol 'system and network administrators' what's the link ?
@skull did what I suggested.
20:43
@TedShifrin Oh okay just dumb the url in the 'link' part...
haha
Right ...
@KajHansen Nevermind technical difficulties here...as you were...
Ah, this makes sense now I think.
You encountered that tacitly all over your crummy alg text, @Kaj.
Lets try this again...Can someone check out the following post...It does not require any PDE knowledge...it's an application of the chain rule I think...check the question by Alex [question about post using the chain rule]{math.stackexchange.com/questions/652214/…}
20:46
You aren't referring to your algebra text, are you, @TedShifrin? :)
@Lucio ... Parentheses for the link.
If the foo shits, @Kaj.
@Will: I would expect you are related. Same for me and Shifrins ...
It works! Now to sort out the maths...
I answered, @Lucio. Are you also Alex?
Heya
Prof. @Ted
Heya @Studentmath!
20:52
@r9m Be around ... I wanna show you another thing
I think I am going to fail. I just don't manage the tests.
r9m
r9m
@Chris'ssis Ninja stays alert :)
I've no idea what to do
@Studentmath sleep, stay alert and determined the day of the exam
20:54
Stressing out worrying does not work, @Studentmath. Understanding as much as you can without memorizing is all you can do.
@TedShifrin I will plead the 5th on that one...does the 5th hold on MSE?
It is very hard not to stress about work Ted...Very hard...it doesn't help but it also can't be helped...
True @Lucio, and thanks @G.T.R @Ted, will try to manage as many tests as I can before the 30th..
I've found in 40 years of teaching that students who keep thinking about the grades they need almost always doom themselves.
20:56
@Ted I seriously have to stop thinking about that, too
@Enjoys thanks!
LOL @EnjoysMath
r9m
r9m
@EnjoysMath great advice :P
Or have sex
It will put you to sleep if you're male
@Studentmath Do what you need to reduce stress...just use your time wisely....
@Hippalectryon this is the site pal
20:58
You actually need some stress at the test day, it gets you working faster
I hate those papers when you look at the first page and you say...ef...maybe the second page is better...effin ef...third page....god help me...back to first page....okay at least its just ef...it's a three page test by the way
@Hippalectryon before you ask a question like that, make sure to gather a lot of information first
@Lucio that's not helpful, no..
@skullpatrol I have nearly none

 Root Access

For all you Super Users out there. You have backups, right?
21:01
haha sorry...good luck...
Thanks :P
@Hippalectryon ask what you need in their chatroom
@Hippa you still have a working laptop?
@Studentmath Kind of a miracles, but yes
God is on your side it seems
21:03
@Hippalectryon can you hear the fan blowing very loud ?
@G.T.R No, it's as loud as usual (it's a laptop)
r9m
r9m
@Chris'ssis :D
@Hippalectryon that's where the anomaly lies. My laptop gets noticeably loud when stressed or heated.
@Hippa: Your Pascal's Triangle question made it to top billing on the SE weekly newsletter !!
@TedShifrin :o
@G.T.R Well i can't tell for sure if it's loud or normal
r9m
r9m
21:06
@Chris'ssis have you decided on a book title ? :D
Can you hear something blowing ?
@r9m 300 spartan integrals and series (or something like that) :-)))))))))))
3
r9m
r9m
@Chris'ssis HAHAHAHA :D Awesome title .. I like it :D
@G.T.R Salut, tu aurais une idée sur cet exercice : On suppose que n est la somme des carrés de trois rationnels. Montrer que n est la somme des carrés de trois entiers ?
21:10
@Nico à première vue, est ce que les 3 rationnels sont en fait les 3 entiers qu'on recherche ?
@Nico constat trivial: si deux des rationnels sont des entiers, alors le 3ème est aussi un entier
@G.T.R Je ne pense pas..oui ça c'est vrai, si on part sur deux rationnels géométriquement on est sur un cercle
pourquoi nous parlons Francais
Pourquoi pas ?
Le théorème de Davenport-Cassels est un résultat sur les représentations rationnelles ou entières des formes quadratiques à coefficients entiers. Il est plus connu pour son corollaire en arithmétique concernant les entiers s'écrivant comme somme de deux carrés ou trois carrés. Énoncé général {{Théorème|Théorème de Davenport-Cassels|Soit \scriptstyle q une forme quadratique à coefficients entiers de dimension \scriptstyle n telle que \scriptstyle q(x_1, \cdots, x_n)=\sum_{i,j}a_{ij}x_ix_j avec pour tout \scriptstyle i et \scriptstyle j, \scriptstyle a_{ij}=a_{ji}. Si, pour tout n-uplet \...
Oui c'était vraiment pas trivial..
When someone says the last few years' tests are relevant - how far does it usually go.. 5 years?
21:21
lol, encore un truc de l'ens ? @Nico
oui
lol
pfff
oui c'est n'importe quoi
I have no clue, @Studentmath, but you get an idea of the style of questions and what topics are covered. Few usually suggests 2 to 4?
Ach so, will go over 2011, they have solutions.
The question seems to be quickly solved once you get the path, 1 hour for all the 4 should suffice if you get right away the path to solution. So I guess he takes in mind 2 hours of trying to figure out how to do it..
21:29
3 hours for 4 questions?
Yep..
Most still don't make it in time.
prelim?
So that means you can stay calm and take time to think, rather than believing you have to know the answer when you start.
It's 3 hours for 4-5 questions in all the math tests in the uni, but some questions take a full hour just to write down the proof
Yes, in this test that is so
He plans on us thinking, just need to be calm and hope I will figure it out eventually
@Tom: Undergrad in Israel
21:31
Oh, I never had an undergrad exam last 4 hours...
Calm is important, @Studentmath. You're smart. Believe in yourself.
Thanks Prof. @Ted!
Still got 4 days to learn, that's a lot. Anyhow, back to it studying - good night all!
Night :)
21:32
And be sure to get some sleep... very important
@Ted do you know what Did has in mind here math.stackexchange.com/questions/845533/… ?
I went into a very important exam once on 4 hours sleep, and it was a disaster. I kicked myself for weeks after because the questions were very doable.
common mistake^
live and learn :-)
yep
next time I took a couple doses of Nyquil
21:40
yep, it's hard to sleep when you're nervous...
the only problem with Dextromethorphan (the thing that makes you sleepy in Benadryl, Nyquil) is that it really makes your mouth dry
c c
c c
some natural way, is just water, drink and put some on your face, it reduces heart rate, and you may sleep
@GTR:Je n'ai aucune idée.
c c
c c
21:57
I don't understand really well why: $\frac\partial\partial{x}(f(tx_1,tx_2,...,tx_n)) = \sum_{i=1}^n x_i\frac\partial{f}\partial{x_i}(tx_1,tx_2,...,tx_n)$
$\frac{\partial}{\partial{t}}(f(tx_1,tx_2,...,tx_n)) = \sum_{i=1}^n x_i\frac{\partial{f}}{\partial{x_i}}(tx_1,tx_2,...,tx_n)$
You have two functions. One is $\varphi:\Bbb R\times \Bbb R^n\to\Bbb R^n$ with $\varphi(t,{\bf x})=t\bf x$.
Is some $f:\Bbb R^n\to \text{somewhere}$.
You're considering $f\circ \varphi$.
@cc chain rule
You want $\partial x$ or $\partial t$?
$\partial t$
c c
c c
$\frac{\partial{g o f}{\partial{x_i}}}=\sum_{i=1}^{n}\frac{\partial{g}{\partial{y_i}}\frac{\partial{‌​f_i}{\partial{x_i}}}$
gives up, yep ok chain rule, or composition rule, thanks
22:13
Heya mr @Pedro ... This looks a lot like Euler's Thm @cc
@TedShifrin Hello.
c c
c c
22:29
@TedShifrin Exactly, Euler's identity
of homogeneous functions
@cc Did you get it, then?
c c
c c
@PedroTamaroff yes, no problem, with your $\varphi : t \to tx$
I'm off, seeya
later pal
22:58
I was wondering if we could make the set of irreps of a compact group G a measure space so that some kind of character orthogonality relations would hold. came across this question, which seems very nice, except now the date on which it was posted has me paranoid I'm a fool.
@blue lulz
if I have a function whose graph I think is a circle, is there a good (numerical) way of finding the center and thus radius?
@MikeMiller You can find the center by drawing two chords, bisecting them and fining the intersection of the perpendicular lines to them.
In fact you don't even need the circle to be complete, you just need an arc of it.
ok, let's say I have a function $g$ such that $g(t), 0 \leq t \leq 2\pi$ draws out the circle
how do i find the center (in terms of stuff I do to $g$ - no geometry, I need to plug in numbers and do stuff to the output)
@MikeMiller Dunno bro.
23:10
alas
Maybe some integration?
The center is the center of gravity.
center of mass?
Yes that.
unfortunately I don't think I can do that without being able to integrate over the interior
and there's no reason as far as I can tell that this circle should be swept out "evenly"
in fact it's not
@MikeMiller Cannot you even out the sweeping?
23:17
how
There is the arclength parametrization right?
oh god
This gives even sweeping. Ask @TedShifrin
I'll try that
don't ping him, he'll make fun of me
anyway I'm not convinced it would help, since like I said, how do you find center of mass if you only have the boundary
Could you explain your setup precisely?
23:22
I have a function $g(t)$
I think the image of $g$ is a circle in the plane
If it is a circle, I want a method that gives me the center
(if it isn't it'll be easy to see when I test to see if everything is equidistant from the center)
@MikeMiller Is this in the complex plane?
@MikeMiller Then the quantity $g(t) + ig'(t)$ is constant and equals the center of the circle.
ah, and I can get the latter numerically
thanks!
@MikeMiller No problem :)
23:25
do you have a reference for that?
probably easy to see myself, but I'm lazy
I thought you wanted to find the center of mass of the interior of a closed curve using line integrals of a parametrization of the curve
@AntonioVargas That's only for arc-length parametrisation, isn't it?
@blue I want what I said up above
@DanielFischer Oh right, I did assume that.
hmm
(My main goal is that I don't need to do a whole lot of on-paper calculation: I want to be able to get a machine to do the numerics without doing reparameterizations by hand, etc)
I suppose putting this another way: three points determine a circle. if I have the three points, how can I find the center?
23:31
why can't you use the chord/bisector thing pedro suggested? wouldn't that lead to a system of equations with a unique solution? you could pick 3 random points from $g(t)$, write the equations of some chords, write equations for the bisectors, etc.
also hi :)
maybe I misunderstand the problem
I think you're probably right, that's likely the best method
COUGH COUGH COUGH
23:36
@Mike: The three points is simple. You find the intersection of the angle bisectors of the triangle. But I think we can do your original problem.
@Pedro: You dying?
Hi, @AndrewG !
heyo @Ted :)
Nah, I suggested the method some time ago.
just finished a history class. Free to play with math! woo
I've been converted to a new point of view, @Ted: "This piece of junk probably doesn't draw a circle"
@Mike: If you integrate $\dfrac 1L\int_C \vec x \, ds$ (arclength parametrization), you get the center of mass of the curve.
23:38
can one numerically arclength reparameterize?
We can get that for an arbitrary parametrization by integrating $\int_a^b \vec g(t)\|\vec g'(t)\|\, dt$.
You need to take diff geo and learn how to use chain rule instead of ever reparametrizing :P
I don't think you comprehend how little I would like to see any of what I'm doing right now ever again for the rest of eternity
What @Antonio said seems quite wrong to me.
23:40
no offense, of course
I've said that about most of my career, @Mike.
@TedShifrin Which one? I say lots of wrong things.
LOL :)
of course, this is just a test case in the first place
That $g(t)+ig'(t)$ gives the center. That's just no good.
23:41
I have to find the damn circle in general
@TedShifrin Hah, yeah, well it is if you're parameterized in a civil manner.
Well, that's the whole problem.
Anyway, the only justification I had for it being a circle is that when I draw it it looks like one
@Pedro: Did you sort out that combinatorics? I rather liked that exercise :)
Balderdash @Mike.
@TedShifrin Not really, no.
I did sort out another one, though.
23:43
But even getting the right "center" doesn't mean it was a circle, @Mike. You need to then show that the radius is constant.
@Ted Of course! That was the next step... checking about a thousand points and seeing if their "radius" was negligibly different.
@Pedro: It's funny. I'm not good at discrete math stuff, but you're ordinarily infinitely better at it than I am ...
It could still be an ellipse of negligible eccentricity, @Mike, I suppose.
@Ted What happens to circles when you take the complex square root? You get an ellipse, right?
You mean a circle centered at an arbitrary point? What do you mean by square root?
@TedShifrin The principle branch, a circle that doesn't touch the negative real axis.
23:47
Oh, so the origin is not inside the circle.
Hmm, I dunno.
I think it goes to an ellipse. But damn, this is turning painful.
I'd have to work on it a while. I've never seen this before.
I think no.
23:52
I suppose I could be lazy and use Mathematica to draw pictures :P
That's exactly what I did, hah
If a circle is described by $|z-a|=b$ then its image under the square root should be $|z^2-a| = b$.
In general that's not going to be an ellipse.
Definitely NOT an ellipse :P
Garbage. All garbage.
Pretty garbage.
23:58
Why, thank you, @Mike. You're so sweet.
I know you are but what am I?
You guys aren't garbage. The fact that the damn thing isn't an ellipse is :P
Sigh. Now the rest of this is hopeless.

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