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9:00 PM
@TedShifrin What's your opinion on atonal/serialist music (basically a good chunk of 20th century music)?
 
not my favorite, @Clarinet. My dad's was tonal, as far as I can tell. :)
 
@TedShifrin Have you ever heard the classic Threnody for the Victims of Hiroshima by Penderecki?
 
r9m
@Chris'ssis I've never tried ignoring anyone b4 ... I will try that sometime just to see how the chatroom looks like on ignore :P
 
@TedShifrin I did this, as $z_0$ is an isolated zero there exist an integer $k$ such that $F(z)^{k}\ne 0$, noted $p$. So I have $F(z)=(z-z_0)^p\sum_{k=p}^\infty F^{(k)}(z_0)/k! (z-z_0)^{k-p}$.
 
Perhaps, @Clarinet. Not sure.
 
9:01 PM
@r9m hahahahaha :-))))))))))))) MEGA STAR
 
You mean $F^{(k)}(z_0)$, @Gato?
 
@TedShifrin Yep :).
 
@TedShifrin si je suppose que f est surjective comme F est de dimension finie alors je peux poser $\{f_1,...,f_n\}$ une base de F comme f est surjective alors il existe $\{e_1,...,e_n\}$ tel que $f(e_i)=f_i \forall i\in\{1,...,n\}$
 
You're making it unnecessarily complicated. Just write $F(z) = \sum_{k=0}^\infty a_k(z-z_0)^k$. Pick the first nonzero $a_k$. :)
Oui, très bien, @Vrouvrou. Et puis?
 
9:03 PM
@r9m you very rarely get annoyed! That's an amazing performance! I wish I could reach this level of yours! :-)
 
@TedShifrin much better. Thanks
 
hi everbody
 
Hello @Lucas
 
r9m
@Chris'ssis haha! I'd take that as a complement! :P (but if I could get annoyed at times I feel I'd have more friends :P)
 
@TedShifrin je considère G le sous espace de E engendré par $\{e_1,...,e_n\}$ et on a $f:G\rightarrow F$
et je n'ai plus d'idées
 
9:07 PM
Il faut considérer un ouvert quelconque.
 
@r9m Yeah, I mean this is a really great trait, not just a complement! I'm not like that unfortunately. :-)
 
@TedShifrin mais comment ?
 
how can I figure out without using formula $cos(a-b)$ that $cos(\pi - x)=-cos(x)$, I mean at the sign, why is " - " and not " + "
?
or $cos(\frac{\pi}{2}+x)=-sin(x)$
how can I understand when is " + " and when is " - " ?
 
Draw a triangle. Or a unit circle.
 
keyword : symmetries of the circle
 
9:11 PM
Or the graphs.
 
@TedShifrin je considére un ouvert de G que je note U , comment utiliser la propriété de f pour dire que f(U) est un ouvert ?
 
Chaque ouvert se reconnaît comme $U+x_0$, où $0\in U$.
 
??
pourquoi ?
 
$B(x_0,\epsilon) = x_0 + B(0,\epsilon)$, n'est-ce pas?
 
oui
 
9:16 PM
Et, donc, si on regarde l'application $f$ ?
 
@r9m by the way, I didn't ask you how you would appreciate (assess) the difficulty of the Knuth's problem on a scale from 1 to 10. :-)
 
@TedShifrin aucune idée
 
r9m
@Chris'ssis idk! getting to the integral .. 5/10 and solving the integral 7/10! so i'd grade that a 7! :P
 
Pourquoi $f(B(0,\epsilon))$ est-ce ouvert?
 
@r9m I also think to somehow generalize his problem in terms of higher powers of the logarithm in the left side.
 
r9m
9:19 PM
@Chris'ssis that'd be awesome!
 
@r9m Indeed!
 
@TedShifrin c'est lauestion sur laquelle je bloque
 
Mais il faut employer ce que t'as remarqué à propos de $G$.
 
G est de dimension finie
 
Oui, mais $f(B(0,\epsilon)) = f(G\cap B(0,\epsilon))$. Et enfin?
 
9:21 PM
@r9m Knuth is 77 years old and he still proposes interesting problems, he's very active.
 
franchement je ne vois pas
 
r9m
@Chris'ssis I am a great fan of his! :-)
 
@r9m hehe :-)
 
Une application en dimensions finies qui est surjective est ouverte aussi.
 
Someone want to help me with this?
 
9:23 PM
on veut le démontrer @TedShifrin
 
He wants to know why $\cos(\pi-\theta)=\cos\theta$ and similarly for sine.
 
Mais fais-le directement, en sachant que $f(e_i)=f_i$, etc.
 
lol at "but then I am not RH Bing"
 
@Balarka: Are you pulling another all-nighter?
 
nah, just going to sleep.
 
9:25 PM
@TedShifrin je ne voix toujours pas
 
ooh, Morandi talks about a bit of galois cohomology. maybe I should read this stuff alongside mult. calc. this summer.
 
Si on regarde $f$ comme application $G\to F$, c'est un homéomorphisme.
I knew you wouldn't last in multivariable calc/analysis, @Balarka :P
 
i knew you would say that, haha
 
BTW, @r9m, let me know how it goes ignoring me :P
 
@TedShifrin Ted! How are you?
:D
 
9:28 PM
don't blame me : i just keep looking at high-falutin' stuff. it's my habit. i am just gonna do mult. calc/analysis this summer, don't worry.
 
heyas @Stan
I'll be sending you exams, @Balarka :P
 
r9m
@TedShifrin I haven't thought out whom I should put on ignore :P (if I do it'd be temporary only) lol
 
sure thing.
 
I'm about to make granola :)
 
They're too easy, though. :(
Thanks for telling us, @Stan :P
 
9:30 PM
ok, i am going to bed
 
@TedShifrin No problem. I'll give you the play by play as I make it, just because I know you find it soooooo entertaining. :P
 
g'night y'all
 
Night, @Balarka.
 
good #timeofday
 
LOL ... appropriate for Mike.
 
9:31 PM
Hello q_q
 
Je ne te réponds plus, @Hippa ...
 
And I think I made an error. I meant to say $\mathbf{F}(c,\mathbf{x})$. I think that should make sense, yes?
@TedShifrin
 
it makes more sense, @Stan, but still missing a variable.
 
lambda
 
Yuppers.
Now count dimensions and see if things make sense, for starters.
 
9:33 PM
Well, n + 2 right? Because it's the number for $\mathbf{x}$ which is $n$ and then you have one for lambda and one for c.
 
OK, and mapping to what?
 
no, no, no, $\Bbb R^?$ ?
 
Um, n?
 
What are the equations for LM?
 
9:36 PM
I've got a projector $p$ of rank $r$ on a finite dimension vector space $E$ and $\phi(f\in\mathcal{L}(E))=(p\circ f+f\circ p)/2$. The eigenvalues of $\phi$ are $0,1/2,1$, how do I get the eigenspaces ? I tried constructiong them (for instance, $E_0(\phi)$ would be imo the set of functions $f$ such that $f(\ker\phi)\subset\ker(\phi)$ and $f(Im\Phi)\subset\ker\Phi)$ but when I add their dimension I don't get $n=\dim E$.
 
$L = f(\mathbf{x}) - \lambda ( g(\mathbf{x}) - c) $
$\nabla f(\mathbf{x^*}) = \lambda^* \nabla g(\mathbf{x^*})$
 
So, how many equations, @Stan?
 
$n$ and then the constraint equation, so $n + 1$
 
so we're mapping $\Bbb R^{n+2}\to\Bbb R^{n+1}$.
So, with appropriate derivative rank condition, we expect a one-dimensional manifold for $\mathbf F=\mathbf 0$.
 
That's awesome. I didn't get that before. Cool! So then, in your example with $\mathbf{F}(\mathbf{x},\mathbf{y}) = \mathbf{0}$ does the 0 vector here have dimension $n+1$?
 
9:41 PM
@r9m btw, thank you for giving me the Knuth's problem, I would have missed such a beauty! Unfortunately I'm not aware when such problems are out (or I find them after a long period of time).
 
So there's one "free variable," which we expect will be $c$ ?
Sure, @Stan.
 
Well, that variable is just the constraint restated right?
 
@Chris'ssis o/
 
@TedShifrin si je prend B(0,1) pour dire que f(B(0,1)) est ouvert il faut qu'elle contienne une boule ouverte qui contient 0 comment je fait svp
 
What do you mean by "free"?
 
9:43 PM
@Hippalectryon \o/ :-)
 
No, @Stan, we're trying to claim that the (optimal) solution $(\mathbf x,\lambda)$ varies smoothly with our budget $c$.
 
r9m
@Chris'ssis you can check Tauraso's page! he posts solutions to old problems that he has attended as well! :D (he keeps updating the list as he solves newer problems ... he puts the solution pdf once the last date of submission is past though .. for obvious reasons :P)
 
Yes, I follow that. Did I say something inconsistent with that?
 
@r9m is the guy also here on MSE? I didn't look at his problems and solutions. I need to find some time for that.
 
@Vrouvrou: là-bas
 
r9m
9:45 PM
@Chris'ssis I don't know! he might or might not be .. maybe uses an alias?(in case he's here)
 
Sorta @Stan.
 
je n'ai toujours pas compris @TedShifrin
 
How so?
 
@Vrouvrou, vraiment? $f|G$ et $(f|G)^{-1}$ sont toutes les deux continues.
 
@r9m Don't miss 11828 on his page. It's interesting.
 
9:47 PM
"Well, that variable is just the constraint restated, right?" :D @Stan I'm saying it's what parametrizes everything.
 
Yes, I knew what I said wasn't right, but I didn't have the words / understanding to state it that well. So I said something subpar and figured we'd end up correcting it. Yay, now I know. :)
 
Did you end up finishing part (a) correctly, @Stan?
 
r9m
@Chris'ssis ah! that also is giving me neck pain! I dunno how to deal with it ... first I thought of faa de Bruno's formula .. but couldn't make it useful though ..
 
Yes, but I have to type it up.
 
OK.
How are your actual math classes going?
 
9:49 PM
I will do that today and send it to you so that we can be sure I did it correctly :P I am going to take real analysis in the fall. I had to wait because I resumed school spring quarter and haven't taken placement tests.
 
@r9m I didn't ponder over it, but I'll do it at some time.
 
It probably doesn't start 'til fall, anyhow.
 
Otherwise, I would have to take remedial calc lololol
which here is pretty low
 
r9m
@Chris'ssis okay! :) also see .. 11821 it looks like a nice limit problem :)
 
That's surprising, since their Honors calc is looney tunes.
Well, the Spivak year is fine. But the next year ...
 
9:50 PM
@r9m Oh, yeah. It's not bad. :-)
 
Hmm...well, I dunno. I certainly don't think it seems as intense as the Calc I took at Caltech. and my brother took the intro calc sequence and he found it easy and he's terrible at math.
Sorry, he took the u of c sequence.
 
LOL, maybe not so terrible as he thinks.
CalTech still using Apostol?
 
Yup, I have become less enchanted with it overtime, but I still like it.
 
Well, I prefer Spivak + my book :) But I'm very biased. Apostol is good, though, just not as good. :)
 
I really like your teaching in general. I haven't used spivak yet. Do you recommend him? The problem is, one of the books I went looking for (diffeo geo maybe?), he had like 5 volumes and that was an immediate turn off because I don't think I can read 5 volumes
lol
 
9:53 PM
I was referring to his Calculus. The 5-volume book is good, but too drawn-out scrambling and unscrambling notation. But plenty of great stuff in there.
 
Oh!!!! I have looked through his calculus book now that I remember. Yes, it's a decent book. I liked it.
 
Is $\{(a, b), (b, a), (b, b)\}$ anti-symmetric? I thought it wasn't because $(a, a)$ isn't in the set, but $(b, b) = (a, a)$
 
The first volume has all the basics on manifolds. Volume 2 gets a bit much with four different definitions of connections on Riemannian manifolds, but Volume 3 has some wonderful concrete stuff in it.
Why is $(b,b)=(a,a)$?
 
Hello friends.
 
@AlexW :)
 
9:55 PM
@TedShifrin!!
 
@r9m I'm done with it.
 
How's it going?
 
About to go cook dinner, @AlexW ... how're you?
 
r9m
@Chris'ssis wow! how? hint please?! :)
 
I love manifolds. But economists don't seem to talk about them much.
@TedShifrin
 
9:56 PM
Not at the undergraduate level, no, @Stan; they can't. But if you go to a high-powered grad program ...
 
Why not?
 
@r9m I cleverly used integral representation of the gamma function. :-)
 
Because the average econ undergrad barely knows single-variable calculus.
 
Not bad, @TedS. I managed to walk to get my groceries before it thunderstormed out. And now it's sunny again, so I may go for a walk soon. :) What's for dinner?
 
Oh......well, yes in that case that would be hard
 
r9m
9:57 PM
@Chris'ssis oho! wait gimmie some time to think about it then :D
 
@Hippalectryon ça va ?
 
@r9m OK :-)
 
I think I'm going to make my once-a-year shot at mac 'n cheese, @AlexW :)
 
@Gato Oui pour l'instant :-)
 
@TedShifrin Because if aRb and bRa, then a = b
 
9:57 PM
pour l'instant ?
 
@Gato Les résultats d'admissibilité ne sont pas encore sortis
 
@Hippalectryon C'était prévu pour ajd ?
 
Ooh, that sounds exciting @TedS. I do love good mac 'n cheese. :D Do you do anything crazy, or are you pretty classical?
 
18h ?
 
(I consider classical strictly noodles and cheese, though the kind of dough/noodles and cheese may vary...)
 
9:58 PM
Oh, that's the definition of skew-symmetric? @Don
 
@Gato Non non, c'est fin mai/début juin
 
je me disais bien, tu as le temps pour stresser.
 
Would you say that that graph is $f\left(x\right)=\frac{1}{2}^{-x}$
 
Tu révises les oraux ?
 
9:59 PM
@TedShifrin Okay, I'm off to make granola. I will email you señor. ¡Hasta luego!
 
@Don: Then it seems to me that if $a\ne b$, we shouldn't have $(a,b)$ and $(b,a)$ both in there.
LOL, bubye, @Stan.
 
@Hippalectryon Je me disais bien, tu as le temps pour stresser. Tu révises les oraux ?
 
Simplify that and answer it yourself, @Maximilian.
 
Simplify it?
How?
 
@Gato Oui
 
10:00 PM
Use exponent rules.
 
so y=2^x?
 
@TedShifrin anti-symmetric: if aRb and bRa then aRa
 
@TedShifrin Do you have any good exercice in complex analysis ? (pour m'entrainer pour mercredi..)
 
@AlexW: I may throw in some bacon ... :P
 
10:00 PM
but since bRa and aRb and bRb isn't in there it can't be anti-symmetric.
 
@Gato: If you email me, I can email you my midterm and final from when I taught the grad course.
 
Oh my. You've got good taste, @TedS. :)
 
@tedshifrin y=2^x is correct? matches the picture?
 
You guys can come to SD and get dinner sometime, @AlexW :P
yes, @Maximilian. Precisely :)
 
OK... cool
The instructions the teacher gave kinda confused me:P
 
10:03 PM
@TedShifrin still the ugh.edu adress ?
 
That would be great @TedS, I'll badger Mike about it. Btw, what have you against Rachmaninoff? ;)
 
uga.edu, @Gato
I just find him overly schmaltzy and uninteresting, @AlexW :)
Second-rate :P
 
I cannot get these graphs:P
 
OK, I'm going to cook. Y'all misbehave without me.
 
Haha, I see @TedS. I enjoy him, but I certainly see that point of view. One of my friends shares that sentiment rather intensely. He tends to enjoy more modern composers, ala Ives, Bartok, etc.
Enjoy dinner, @TedS!
 
10:05 PM
Can anyone explain relative topology?
 
@TedShifrin Done :D.
@Hippalectryon Tu ne parles plus avec Ted ?
 
@Gato Uh pas en ce moment :c c'est compliqué
 
Would be $f\left(x\right)=2^{\left(x-1\right)}$
 
@Hippalectryon why ?
 
correct?
 
10:09 PM
@Gato Tu te rappelle du meme que j'avai fait sur Ted d'après sa Vidéo ?
@Gato On a eu récemment un clair désaccord sur qqch lié, d'où la situation actuelle
 
@Hippalectryon même ?
 
J'ai déjà vu cette photo, c'était de toi ? ^^
 
Oui oui, l'année dernière
 
Et donc Ted n'a pas aimé ?
 
10:20 PM
@Gato Ted connais cette image depuis longtemps, mais on a eu un désaccord sur je ne sais plus trop quoi lié à ça, qui a mené à la situation actuelle
Bon, il est minuit 20 :c je devrais y aller
 
D'accord, faut régler ça donc. Pareil, bonne nuit!
 
Bonne nuit :-)
 
@tedshifrin is the inverse function of a log just the negative version of the exponent form?
E.G. $y=log_x(x-1)$ inverse is $x=-(2^y+1)$?
 
0
Q: What is the probability that three randomly chosen vertices of a regular 100-gon inscribed in a circle form a right angled triangle?

sourishA regular polygon with $100$ sides is inscribed in a circle. What is the probability that three randomly chosen vertices of this polygon form a right angled triangle?

In the answer I wanted help in finding out how he got $50.98$.
 
10:36 PM
there are 50 diameters (# of points divided by 2) and there are 98 choices for the third point
also use \cdot instead of a period to denote multiplication
 
r9m
@Chris'ssis it's 4 am here .. I'll get some sleep :) there's a nasty expression I'm stuck with .. with a bunch of $p^{th}$ derivative and stuff .. I need a fresh start .. g'night!
 
@r9m OK :-) Good night! :-)
 
10:59 PM
@anon Thank you Sir for your help and I will keep in mind to use \cdot in future.
 
11:30 PM
$\int \frac{r^{3}}{\sqrt{r+r^{2}}}*dr$ = $\int \frac{r^{2}}{1} * \frac{r}{\sqrt{4+r^{2}}} * dr$.
u = $r^{2}$, du = $2r$, dv = $\frac{r}{\sqrt{4+r^{2}}}$, v = $\sqrt{4+r^{2}}$
$ = (r^{2})(\sqrt{4+r^{2}}) - \int{ (\sqrt{4+r^{2}})(2r)}*dr$
$ = (r^{2})(\sqrt{4+r^{2}}) - 2\int{ (\sqrt{4+r^{2}})(r)}*dr$
$ = (r^{2})(\sqrt{4+r^{2}}) - 2\int{ (\sqrt{u})(r)}*\frac{du}{2r}$
$ = (r^{2})(\sqrt{4+r^{2}}) - \int{ (\sqrt{u})}*du$
$ = (r^{2})(\sqrt{4+r^{2}}) - (\sqrt{4+r^{2}})^{3} * \frac{2}{3}$
Somehow this is wrong, can someone see where I messed up?
pls
 
11:48 PM
was your original integrand $\frac{r^3}{\sqrt{r+r^2}}$ or was it $\frac{r^3}{\sqrt{4+r^2}}$? you wrote both in your first line.
 
4 + r^2
Just a mistake
 
$\displaystyle\int\frac{r^3}{\sqrt{4+r^2}}dr=r^2\sqrt{4+r^2}-\frac{2}{3}(\sqrt{4‌​+r^2})^3+C$ is correct
why do you think it's wrong?
 
Wolfram Alpha keeps giving me something else.
 
what does it give you?
 
(When I try plugging in numbers)
 
11:51 PM
explain
 
It is a definite.
Between 0 and 1.
So I plugged both 0 and 1 in.
And subtracted the result of plugging (1) in from (0)
 
and then what?
 
I don't get what Wolfram Alpha gets.
 
what do you get?
 
Erased from my calculator, let me input again
(1sec)
-4.43
Wolfram gets 0.115811
 
11:56 PM
then you typed your stuff into the calculator wrong
you shouldn't be using a calculator anyway
plug things in by hand and then tell me what you get
 

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