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12:17 AM
@Bitrex There's a 60 year old guy doing his PhD in my office
 
@DanielRust Cool!
 
12:39 AM
@Bitrex: i've certainly had students older than I. Maybe 50's. how old be you?
Older than I was, I meant.
 
@TedShifrin 35 as of the 15th :)
 
Oh, you're still a young'un. What level of school? :)
 
@TedShifrin I'd like to take some university level math courses; never done that before. I don't have any formal math education outside of high school.
 
Wow, awesome. You have a college/uni nearby?
 
@TedShifrin I'm right outside Boston.
 
12:58 AM
Oh, zillions of colleges. That's my old home ... 10 years' worth.
I'm sure Northeastern has lots of non-traditional students, for example.
 
1:35 AM
@TedShifrin Harvard Extension School gave a course in abstract algebra a while back, unfortunately at the time I was too ill to attend it.
 
I'm sure you'll have lots of options ... @Bitrex.
 
@TedShifrin Coursera's also got an online course in complex analysis coming up October 21st. Looks pretty good!
 
Personally, I'm not fond of all this new-fangled on-line learning, but I know lots of folks are.
2
 
1:55 AM
Does anyone understand this question? math.stackexchange.com/questions/496134/…
 
 
2 hours later…
4:15 AM
@MarianoSuárez-Alvarez whats yellow and equivalent to AC?
 
4:48 AM
@PeterTamaroff Get bit by something?
 
5:33 AM
@KevinDriscoll Dude, links. Use them.
 
Oh dayum. Burn.
Didn't realize it was a link.
 
5:57 AM
@PeterTamaroff No tenes que ir a dormir?
 
@Bill Daf...? Do I know you?
 
Maybe
Guillermo Tochi
 
@Bill Oh.
Duh. "Bill".
@Bill Creo que podemos hablar en castellano.
 
Estoy con fer aca
 
@Bill No los encontré hoy. =/
 
6:00 AM
No, la clase de fer se cancelo
 
Habia un seminario o algo.
 
porque habia seminario de teoria de numeros
 
@Bill Si, eso!
Supongo que no van a dormir?
 
Y, no da
pensabamos alcoholizarnos mientras estudiamos equivalencia morita
 
@Bill Jajaja, que es eso?
 
6:02 AM
No te aburris en lineal?
 
@Bill Por?
 
Me da la impresión que ya sabés lo que estan dando
 
@Bill Hmm... no todo, pero bastante si.
 
hay un libro de greub
de springer
es mas avanzado
tal vez tenes cosas que no sabes ahi
en la leloir esta
 
@Bill Ah! Dos segundos.
 
6:05 AM
si, ese
 
@Bill Maxi nos contó un poco sobre espacios cocientes, conozco las construcciones en anillos y grupos, pero ver cosas "concretas" está bueno, como $V/S\simeq W$ si $W\oplus S=V$.
O que $V\simeq K^{(B)}$ y $V^\ast\simeq K^{B}$ si $B$ es base, en general. Esos detallecitos (nada extraños) no los tiene el libro de Sabia, Tesauri y Jeronimo. =P
 
Eso si B es finito
En general el dual es mas grande
Si, cocientes en espacios vectoriales no son muy divertidos
 
@Bill Por eso, $K^{(B)}$ es el de sequencias eventualmente nulas.
 
ahhhhh
nos cagaste
 
@Bill JAJAJAJA.
Pero no es notacion usual?
 
6:09 AM
Sisi, leí al revés por algún motivo
 
@Bill Ah, =P
 
vos pensas dormir?
 
@Bill Nah. Duermo en el viaje, espero.
 
Ok, le digo a fer que no te charle
 
@Bill Jajja, de ultima duermo a la tarde. Voy a tener que tomar mas cafe que lo comun, soy medio adicto =D.
 
6:12 AM
Jaja, mirá que a la tarde están los cursos!
 
@Bill Drats. Entonces duermo a la mañana? Proposicion Siempre hay tiempo para dormir.
 
y...
llegamos tipo 10 creo
 
(Excepto si sos ingeniero, o médico.)
 
y nos tendriamos que registrar
 
@Bill Claro.
 
6:14 AM
yo suelo hacer ciclos de mas de 24hs sin dormir
 
Yo me tendría que reinscribir a los cursos.
@Bill Muy seguido? A mi me pasa si me engancho con algo de mate.
Ahi estoy listo.
 
si, es imposible decir "listo, ahora duermo"
 
@Bill Exacto.
Te doy un ejemplo no tan reciente:
Fijate la cantidad de revisiones =P
 
mfw
le pusiste ganas
 
@Bill Yes.
 
6:19 AM
Yo le puse muchas ganas a una pregunta de fer P=0
como se editan los links aca?
 
[alt-text](link)
@Bill Ha! Justo la vi, +1. La respuesta mas corta (y correcta) en la historia de MSE es esta:
 
hahaha
 
@Bill Que dia se vuelven ustedes?
 
El viernes a la noche
 
@Bill Mariano me dijo que hay una cena el Sábado, puede ser?
 
6:26 AM
ni idea
 
@Bill Ah. Yo veo, por el momento tengo reservadas dos noches nada mas en el Freedom, despues no hay lugar.
 
y, a una plaza
 
Heh!
@Bill Que dias tenes Topología vos?
 
Lunes y Miercoles
 
@Bill A la tarde?
 
6:33 AM
Noche
de 7 en adelante
 
Ah. Que tal es el turno noche?
 
una garcha
nah, igual que todos
una garcha
 
JAJAJAJA igual que todos?
 
se
 
El curso de Topología es malo? =/
 
6:42 AM
nah, los alumnos son molestos
(soy guillermo ahora)
 
@Bill Ah! Claro, yo me olvido que vos das el curso!
@Bill Molestos en que sentido?
 
hacen preguntas
$\geq$ 1
 
@Bill Hhahaha. Preguntas aburridas, supongo?
 
nah, que se yo
vienen todos con la misma
y aparentemente voy "muy rapido" cuando doy clase
 
Es raro que alguien este con topologia si falla tan terriblemente en calcular un complemento.
@Bill Se quejan?
 
6:45 AM
entre ellos seguro
 
@Bill Ah. Me parece bueno no hacer las cosas lentas. Hace que las personas se pongan las pilas un poco mas!
 
y ademas no me gusta escribir toda oración que pronuncio en el pizarron
y a cierte gente eso no le gusta
 
@Bill Je, eso me volviería loco. "Queremos mostrar que todo base del $K$ espacio vectorial..." NO!
@Bill Supongo que llega cierta etapa que deja de ser necesario.
 
hay gente para todo
pero bueno, cada uno tiene sus tiempos
 
@Bill Que están viendo ahora?
 
6:51 AM
yo soy un idiota para ciertas cosas
en topo?
creo que iban a empezar con topologia de espacios de funciones
 
@Bill Ah.
@Bill Ven Arzela Ascoli?
(O eso se ve ya en Funcional?)
 
se ve en avanzado
 
@Bill Ah, nice!
 
(pero no con toda su generalidad)
 
@Bill Claro.
Yo solo se el caso particular en $C(\Bbb R)$.
En Rudin hay una demostracion buenísima.
 
7:00 AM
cuál es? la que usás lineales a trozos?
la version mas general es sobre espacios localmente compactos, hausdorff, y sigma compactos
 
@Bill Nope. Usa separabilidad y Bolzano Weiertrass.
@Bill Como es?
Wait, no se que es $\sigma$-compacto.
 
union numerable de comapctos
 
@Bill Ah.
Si querés te cuento la demostración de Rudin, de paso la refresco yo. La leí pocas veces.
 
a ver, dale
 
hola
 
7:13 AM
@Bill OK, tomemos un compacto $K$ en $\Bbb R$. Entonces es separable (dem: tomar $1/n$-redes para $n=1,2,\ldots$). Sea $E$ tal subconjunto numerable y denso en $K$. Sea $f_i$ una familia acotada puntualmente y equicontinua.
Perdon, no.
Para cada $x_i$; $$f_j(x_i)\; j=1,2,3,\ldots$$ esta acotada.
La idea es tomar subsuc. convergentes ahora, pero de forma inteligente.
@Bill La notacion es medio engorrosa, pero aqui vamos.
$f_i(x_1)$ tiene una subsucesion convergente. Rudin la escribe $f_{1,k}$ para tal subsucesion de funciones, entonces $f_{1,k}(x_1)$ es convergente.
 
Te querés quedar con la diagonal?
 
@Bill Yep ;)
Es muy linda la demostracion.
La idea es que obtenemos una subsucesion $f_{n,n}$ de la familia $\{f_n\}$ tal que $f_{n,n}(x_i)$ converge para cada $i$.
Se muestra entonces que esta diagonal converge uniformemente,.
 
claro
 
@Bill Supongo que no tengo que ahondar en detalles con vos. =D
 
con fer puede que si
hahah
nah
 
7:26 AM
@Bill Cuando viajás a Mendoza, los de Via Bariloche sortean vinos. Hay un bingo y todo.
Me pregunto si hacen algo parecido en Chevallier jejeje.
 
te gusta el vino?
a mi no
sigue siendo muy de viejo para mi
 
@Bill Nah, a mi tampoco.
Pero me parecio copado que lo hagan.
 
puede que nos den algun diario
 
@Bill Che, a que hora salen para Retiro? Ya van a ser las 5.
 
a las 5
 
7:40 AM
Ah. OK. Yo me voy a ordenar mis cosas.
Nos vemos en un rato!
 
dale
 
I must be in the wrong chat. I don't understand a thing on my screen...
3
 
user87637
8:24 AM
@robjohn I have come to your rescue.
 
Google offered to translate the page
 
 
1 hour later…
9:37 AM
@robjohn do you understand this?
-2
Q: Prove that the coordinate ring for $V × W$ is isomorphic to $k[V ] ⊗_k k[W ]$.

oofhdearProve that the coordinate ring for $V × W$ is isomorphic to $k[V ] ⊗_k k[W ]$ where $V$ is a Zariski closed set of $\mathbb{A}^n$ and $W$ is a Zariski closed set of $\mathbb{A}^m$. Not sure how to do this some help would be nice.

In particular, these comments:
This is your professor speaking. Obviously, I can't stop you from posting questions from the problem set here but I can not for the life of me understand what is the benefit to you to ask here instead of in office hours! Am I really so bad at explaining?
Also, this particular question is proved carefully in the textbook, in the very section that was assigned for reading. Your other questions (for example, about what the pullback map of coordinate rings looks like) show that you are not understanding the most basic concepts in the course. Why are you not asking the professor? The course is going to get a lot harder, and quickly! Would it be better to drop math 631 and take Math 593 instead?
 
10:02 AM
Hellow Guyz , I have a serious math problem .
Can you help me ?
How can I show that un-corelated random variables are not necessarily indepedent ??
Is there anyone ??
@bitrex
@robjohn
 
10:24 AM
scorchio!
 
hi
 
How is it possible that when I compute the geometric series of a truncated power series p(x) and plug in the value x_1 I get the correct result, but If I begin by evaluating p(x) at x_1 I get a value of p(x) outside the radius of convergence for the geometric series and it fails. The expression for the geometric series of the truncated p(x) is not nice at all but it works.
So if you don't tell the geometric series that the value of the truncated power series will be outside its radius of convergence, you can fool the geometric series?
 
10:50 AM
Does anybody know how can I prove that the set of matrices that can be joined to the identity I forms a normal subgroup of $GL_n(\mathbb{R})$
 
@Carpediem what have you tried?
 
@TobiasKildetoft I am not really sure how to start..
 
@Carpediem have you shown it is a subgroup?
 
@TobiasKildetoft I am having trouble defining the actual set
That would get me going
 
@Carpediem: I assume it's meant to be the path component containing the identity?
i.e. the endpoints of all continuous curves starting at the identity
 
10:58 AM
@Carpediem what do you mean defining the set?
 
@TobiasKildetoft The set of matrices that can be joined by a path to the identity I
 
@Carpediem: you just defined it ;)
 
@Carpediem yes, I know what it means. I was wondering what you meant by having trouble defining it.
 
@TobiasKildetoft No but I mean how to write it using the quantifiers
 
@Carpediem why do you want to do that?
 
11:04 AM
@TobiasKildetoft I thought it would be easier in order to prove that it is a subgroup
 
@Carpediem this is really just the same argument as the one you asked about on the main site
 
11:40 AM
Any mods around? I have just had two comments on an answer removed and I don't get why.
 
Was that answer changed into comments?
 
It was a quite nice answer about the connected component of a topological group. I commented that the OP seemed to be asking about the path-connected component instead, which would make the argument even simpler
 
That's what happened to the question I posted above...
 
11:44 AM
the second comment was in response to the OP asking about this, inquiring whether he really did not see how to do this using his previous question
 
link please
 
0
Q: Connected component of $GL_n(\mathbb{R})$

CarpediemThe set of $n\times n$ matrices can be identified with the space $M_n(\mathbb{R})$. Let $G$ be a subgroup of $GL_n(\mathbb{R})$. I would like to show that the set of matrices that can be joined to the identity $I$ forms a normal subgroup of G.

 
@cyberskull Sorry, not at all well-versed in algebraic geometry. I have no idea. The latter part seems to be an angry instructor who thinks this is a student from their class. I don't know how they could be certain, unless the asker gave their real name.
 
@robjohn are you able to see who removed my comments?
 
@TobiasKildetoft where did they remove your comments?
 
11:51 AM
@robjohn above linked question
(they were on the answer)
 
 
3 hours later…
2:43 PM
hi
 
2:57 PM
@TobiasKildetoft It seems they were flagged and removed by non-mods.
 
 
2 hours later…
5:00 PM
Greetings!
@robjohn I've just created 2 new amazing questions.
 
@Chris'ssis nice
 
@robjohn This one has a closed form $$\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \frac{1}{(2i+1)(2j+1)(i+j+1)}$$
 
user87637
5:19 PM
@robjohn Yeah, considering how often questions seem to be recycled, it could come from anyone or anywhere.
 
user87637
I am very excited. My holy math books will be arriving in a couple of weeks from amazon.
 
@JasperLoy It appears that that question had been asked before. It was probably someone else going through the text, not someone from her class.
@JasperLoy couple of weeks? couldn't you get faster delivery?
 
user87637
@robjohn Well, I am not in the US, and I don't need quick delivery, so I go for the cheapest.
 
user87637
Americans are so lucky... Wish I were American... Maybe in my next life...
 
@JasperLoy Okay. I have no idea what overseas shipping costs.
 
user87637
5:23 PM
@robjohn It is good that they ship for free to my country if I spend US125D at least, but other charges apply.
 
@Chris'ssis I bet $\zeta(3)$ is involved... I have to take my son to the doctor before I have time to work on this.
 
user87637
@robjohn I hope he is OK.
 
@robjohn Right! :-)
 
@Chris'ssis If it is anything like the others of that kind I have worked on, the diagonal, which sums to $\frac78\zeta(3)$ appears in a linear relation to the total sum.
 
@robjohn exactly, that is the answer. :-)
 
5:28 PM
@Chris'ssis I'll look closer when I get back :-)
 
@robjohn you were very fast! :-)
 
@JasperLoy yes, it is a check up
 
@robjohn till then I attend the triple series version.
 
@Chris'ssis eek
 
5:57 PM
What is h in the first answer here: math.stackexchange.com/questions/497416/…
Anyone ?
 
$$\huge \text{GODDAMN!}$$

$$\huge \text{HENNING MAKHOLM BELLOW!}$$

http://i3.kym-cdn.com/photos/images/newsfeed/000/243/561/afc.gif
 
Oh, I'm so brilliant!!!!!! I just did a very nice triple series!
(sorry for bragging)
:-)
 
@Chris'ssis DUDE, I FOUND A DEAL WITH IT WIT HENNING MAKHOLM
 
6:16 PM
hmmm, some double series led me to Nielsen’s polylogarithm functions
 
6:37 PM
in math.stackexchange.com/questions/497327/… , in the answer which computes the equation of the line, where does the 530−195 come from?
 
hi
 
@robjohn Hmm, how many flags are needed to get a comment removed?
 
@TobiasKildetoft I think robjohn is at doctor now.
 
@Chris'ssis ahh, thanks
 
6:52 PM
how can I prove that $(A \Union B)(A \intersect B) = (A\B) \Union (B\A)$
 
Let $G$ be a group (finite if that makes this easier) where all elements have order dividing $2$. Can anyone think of a way to define an extra multiplication on $G$ which distributes over the original operation and such that there is a unit for each element (so the unit need not be the same for each element). And do this without using that $G$ is abelian along the way.
For example, we can take elements $e, g_1,g_2,\dots$, each time taking an element not in the subgroup generated by the previous ones, and require these to be orthogonal idempotents in the new multiplication
but I don't see a way to show that this is well-defined without using that $G$ is abelian
 
A and B are collection
 
@pourjour I assume you are missing a $\setminus$ on the left side?
 
$(A \bigcup{B}) \setminus (A \bigcap B) = (A \setminus B) \bigcup (B \setminus A)$
 
@pourjour the left side still does not make sense
use \setminus for set difference
 
7:02 PM
@TobiasKildetoft yes I forgot it
 
@pourjour so take some element $x$ in the set on the left side. What can you say about $x$?
 
@TobiasKildetoft $x \in A \bigcup B and x \notin A \bigcap B$
 
ok, so let's split into two cases, $x\in A$ and $x\in B$
 
then $x \in A or x \in B$
 
right, so if $x\in A$ what does that tell you?
 
7:04 PM
if x \in A
@TobiasKildetoft here I'm stuck
 
@pourjour so you know that $x\in A$ and $x\not\in A\cap B$
is it possible to have $x\in B$?
 
@TobiasKildetoft yes
 
@pourjour great, so we know that $x\in A\setminus B$
do you see why we must have $x$ in the set on the right then?
ohh, didn't you say no first?
 
the same for if x \in B
@TobiasKildetoft I had a doubt
 
well, do you see that if $x\in A$ and $x\not\in A\cap B$ then $x\not\in B$?
 
7:10 PM
@TobiasKildetoft yeah great
 
so we know that the left side is contained in the right side
now try to do the same starting from the right
(btw, an alternative is to note that both sides consist of precisely those elements are are in precisely one of $A$ and $B$ and not the other)
 
@TobiasKildetoft till now we just proved that $(A \bigcup{B}) \setminus (A \bigcap B) \subset (A \setminus B) \bigcup (B \setminus A)$
 
right
so now we should try to show the inclusion in the other direction
 
ok till now we just used x
is there any other way using collections
A and B union intersection and other stuff
 
not sure what you mean by that
 
7:17 PM
we used $x \in $ but sometimes we use collections and and their own operation like "setminus" to prove such stuff
 
I am still not sure what you mean. Could you give an example?
 
@TobiasKildetoft ok I will just try to learn your way and thanks :D
 
7:36 PM
@leo hey! where are you, man? :D
 
7:58 PM
oi @ian
 
@Charlie oi! Como vai a universidade?
 
@IanMateus bem, bem... nas palavras de um colega "eu sei que não está bem porque no IME não dá pra estar bem"
@IanMateus e tu?
 
@Charlie IME? Qual?
 
@IanMateus IME-USP
 
@Charlie ah, sim, bacana :D
 
8:01 PM
@IanMateus bom. já entrou?
 
@Charlie Achei um livro de geometria bacana da minha escola, são os trabalhos do Euclides e do Arquimedes
@Charlie onde?
 
@IanMateus :)
@IanMateus universidade
 
@Charlie não, eu moro muito longe hehehe
 
@IanMateus ah
 
@Charlie bacharelado?
 
8:04 PM
@IanMateus sim
 
@Charlie o que vocês estão vendo agora? Tem alguma coisa online?
 
@IanMateus não , num tem :/
@IanMateus tu pode ver a grade curricular etc
 
@Charlie eu vou ver agora :D
 
@IanMateus :DDDDDD
 
@Charlie qual semestre?
 
8:08 PM
@IanMateus sexto eu acho XD
 
8:25 PM
@Chris'ssis :D
 
@Charlie hey !!!! How are you? :D
 
@Chris'ssis I'm fine, and you???
 
I'm creating some new questions. :-)
 
@Chris'ssis nice :)
 
@Charlie yeah, it's just fun! :D
 
8:29 PM
@Chris'ssis yup
 
Here is an interesting limit $$\lim_{s\to0} (\zeta{(s-1)}-\zeta{(s)})$$
hmm, robjohn seems to be away.
 
@Chris'ssis :D
 
@Charlie also this one is cute $$\lim_{s\to0} \frac{\zeta{(s-1)}-\zeta{(s)}-5/12}{s}$$
 
@Chris'ssis hmmmmmmmmm
hi @anon
 
brb
 
8:40 PM
hello grumpikins
 
okay :)
@anon :D
 
taking one out of my book
 
@anon oh
@anon i have a question for you
it's deep
 
sure
hmm
 
@anon it's driving me nuts
@anon hmm?
 
8:43 PM
"it's deep"
"hmm"
 
@Charlie I didn't put those things at random! Let me show you something beautiful to compute I've just created.
 
@Chris'ssis okay :D
 
$$\lim_{s\to 0} \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{(k+n)^s}$$
 
@anon I ask soon
@Chris'ssis and how you did it?
@Chris'ssis you tell the saint but not the miracle ;)
 
@Charlie It just came to my mind. Well, it's a matter of practice. As regards the solution, I won't take you the pleasure of thinking of it.
:-)
 
8:45 PM
XD
@anon
@Chris'ssis I think i'll think of it
 
@Charlie
 
@anon $\huge \text {what's your name???}$
 
here I thought I was being narcissistic guessing that was your question
 
@anon :D
@anon I'll die without knowing your name
that's sad
 
some days go by in which my name doesn't even cross my mind once.
it is not an important thing.
 
8:53 PM
it is...
don't come with that Romeo & Juliet speech
 
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