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Fargle
Fargle
9:28 AM
I'm trying to prove that there exists $A \in M_n(\Bbb R)$ such that $Tr(A) = 0$ and $A^2 = I$ if and only if $n$ is even, but I'm struggling to divine why I need that $Tr(A) = 0$. Any ideas?
 
David Wheeler
David Wheeler
If you omit that, doesn't $A = -I$ work?
Think about the minimum polynomial of $A$. It has to be a factor of $x^2 - 1$
 
Pedro Tamaroff
Pedro Tamaroff
Your matrix is diagonalizable and its eigenvalues are -1,1. You cannot sum an odd number of those and get 0.
The possible eigenvalues are...
 
David Wheeler
David Wheeler
...and that should give you both directions
 
Understand
Understand
 
David Wheeler
David Wheeler
?
 
Pedro Tamaroff
Pedro Tamaroff
9:36 AM
That's classified.
 
Understand
Understand
@PedroTamaroff Is some element of it unclassified?
 
David Wheeler
David Wheeler
It's odd they have 1 reputation
 
Understand
Understand
@PedroTamaroff How does the classified thing work? If a bunch of non-moderators know about it, can a moderator talk about it then? Or is the information only allowed to transfer from non-mods
@DavidWheeler You go to 1 rep when banned
@DavidWheeler And get it back when unbanned
 
Fargle
Fargle
Alrighty, that helps a bunch. Thanks guys.
 
David Wheeler
David Wheeler
I guess they ticked somebody off. But if I'm not getting a steak dinner out of the deal, I don't care.
 
Pedro Tamaroff
Pedro Tamaroff
9:40 AM
@Understand As a moderator I cannot disclose information regarding a suspension.
 
David Wheeler
David Wheeler
@PedroTamaroff It's cool. You've got rules.
 
Understand
Understand
@PedroTamaroff Even if it is known already?
By a non mod
 
Pedro Tamaroff
Pedro Tamaroff
Well, I don't know what you mean by that.
 
David Wheeler
David Wheeler
@Understand I think it safe to say if Pedro says he can't tell you, you may infer he won't
 
Understand
Understand
I mean for example everyone knows why Bill D was banned, and thus the moderators could talk about it, so maybe as long as people already know why, you can talk about it
 
David Wheeler
David Wheeler
9:56 AM
It's conceivable that in one case to do so might violate someone's privacy
Or maybe the mods are planning a coup-d'etat. Shrugs
 
Understand
Understand
@DavidWheeler This seems likely
 
David Wheeler
David Wheeler
Math SE tried to trick me yesterday
It gave me an edit to approve that was written by a bot
 
robjohn
robjohn
@DavidWheeler That happens to everyone... it is an audit. People approve bad edits too quickly.
Is anyone else having trouble getting to main?
 
Theorem
Theorem
@robjohn Good morning .
 
robjohn
robjohn
10:12 AM
@Theorem hello. I looked at the question but haven't been able to comment on it.
 
Theorem
Theorem
@robjohn Ok . :)
@robjohn I just made some corrections .
 
robjohn
robjohn
@Theorem okay... I just got main back, so I will look.
 
Theorem
Theorem
@robjohn Thanks
 
Understand
Understand
If I have some $V$ as a 2 dimensional vector space, and $\phi: V\to V$, where some matrix $A$ of $\phi$ has a basis $\{v,v'\}$ and a matrix $B$ with respect to another basis $\{w,w'\}$

and I want to find $X$ for $XAX^{-1} = B$ the $X$ is going to rely on the basis right?
 
robjohn
robjohn
@Understand yes
 
Understand
Understand
10:26 AM
Can I have a hint for finding $X$? It's for an assignment question, so nothing big please(if you are willing)
Is it meaningful to say that $A$ and $B$ are diagonals of one another?
 
robjohn
robjohn
@Understand that doesn't mean anything to me. They are called similar.
 
Understand
Understand
Is it fair to say I am looking at:

$$X\,\begin{bmatrix}V_1 & V_2 \\ V_1' & V_2 '\end{bmatrix}\,X^{-1} = \begin{bmatrix}W_1 & W_2 \\ W_1'& W_2'\end{bmatrix}$$
 
Understand
Understand
10:48 AM
This necessarily diagonalises A to B right? Hence $W_2=0$ and $W_1'=0$?
 
Understand
Understand
11:03 AM
 
robjohn
robjohn
11:33 AM
@Understand sorry it took so long. I was writing it up offline so that I could enter it here. However, I have added it as an answer to the question :-)
All those matrices take a while to type up in LaTeX
 
Understand
Understand
@robjohn Thanks very much!! I thought my question might not be interesting
 
robjohn
robjohn
11:51 AM
@Understand It may be a duplicate, and if so someone will mention that. However, the question is a fundamental Linear Algebra result.
 
Vlad
Vlad
hello
How can I calculate x1, y1, z1 and x2, y2, z1 if I know that the distance between two points A(x1, y1, z1) and B(x2, y2, z2) is 10
?
Was it with adjacency matrix or?
and I also know this identity
(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 = 10^2
 
Vlad
Vlad
12:25 PM
no body knows? :D
 
 
1 hour later…
teadawg1337
teadawg1337
1:26 PM
watches a tumbleweed roll by
Hello @Balarka
 
Balarka Sen
Balarka Sen
hi
 
teadawg1337
teadawg1337
Seen any interesting problems on the main lately?
 
Balarka Sen
Balarka Sen
no, but i haven't been following anything except the algebraic topology tag either.
what have you been studying, then, @teadawg1337
 
teadawg1337
teadawg1337
@Balarka Math-wise?
 
Balarka Sen
Balarka Sen
mhm
 
teadawg1337
teadawg1337
1:40 PM
Polylogs and natural logarithms, and manipulating series thereof
 
Balarka Sen
Balarka Sen
yuck
 
teadawg1337
teadawg1337
It is nasty, I agree, but it'll be that much more satisfying to find the solution
I wish I could go into further details, but this is definitely something I intend on publishing in the future
 
Balarka Sen
Balarka Sen
i find it rather ad-hoc to compute and manipulate series and integrals.
weren't you interested in number theory a few months ago?
 
teadawg1337
teadawg1337
Yes, and I still am, but this proof has been haunting me and I feel compelled to complete it
 
Balarka Sen
Balarka Sen
you should study a teensy bit of complex analysis, if you haven't already, and then read up bits on the Riemann zeta function. it should go well with your interest in analysis.
 
teadawg1337
teadawg1337
1:49 PM
The Basel problem is what sparked my interest in analysis in the first place
The approach I took forced me to learn about polylogs, and I've been fascinated by complex analysis ever since
 
Balarka Sen
Balarka Sen
cool. d'you know that $\zeta(2) = \pi^2/6$ implies infinitude of primes?
 
teadawg1337
teadawg1337
I did not
 
Balarka Sen
Balarka Sen
can you try to prove it?
as a hint, consider the Euler product for $\zeta$.
 
teadawg1337
teadawg1337
Sure, I'll get back to you once I've come up with an answer
 
columbus8myhw
columbus8myhw
Hello!
 
teadawg1337
teadawg1337
1:53 PM
Shouldn't take more than 5 minutes
 
Balarka Sen
Balarka Sen
shh @columbus8myhw ;)
 
columbus8myhw
columbus8myhw
sorry
 
Balarka Sen
Balarka Sen
and no, what you said is not enough.
though close
 
columbus8myhw
columbus8myhw
Is there a way to "whisper" on this chat room?
 
Balarka Sen
Balarka Sen
:P no
 
columbus8myhw
columbus8myhw
1:54 PM
Like, send a message to only one person.
OK.
 
Balarka Sen
Balarka Sen
i know you can do it, no need to convince me
=)
 
columbus8myhw
columbus8myhw
@Exterior Hello
 
Exterior
Exterior
hello
 
teadawg1337
teadawg1337
Euler's product is $\prod_{p\in{P}}\frac{1}{1-\frac{1}{p}}$, correct?
 
Balarka Sen
Balarka Sen
yep
@Exterior thought about the snake map anymore?
 
Exterior
Exterior
1:57 PM
just got back from practice, so not yet
 
Balarka Sen
Balarka Sen
well, it seems complicated.
 
Exterior
Exterior
@me or @tead?
 
columbus8myhw
columbus8myhw
What's the snake map?
 
Balarka Sen
Balarka Sen
@Exterior i'm talking about the snake map
 
Exterior
Exterior
in what sense? obtaining a formal proof?
geometrically it is indeed obvious, I'd say
 
Balarka Sen
Balarka Sen
1:59 PM
that it is zero for the long exact sequence for $(\Sigma_g, A)$
i can't figure out what it does to the rest of the cycle, the bit not in $B$.
 
Exterior
Exterior
it doesn't matter, since anything not in $B$ is (intuitively) made from chains in $A$
which means it's in the same homology class as $A"\in " Z_n(X,A)$
(I might be wrong)
 
Balarka Sen
Balarka Sen
well, yeah, but how does that say it is sent to 0 by $\partial$?
ohh
 
Exterior
Exterior
is that an "ohh" of agreement or of your own insight? :P
 
columbus8myhw
columbus8myhw
May I ask what the snake map is?
 
Balarka Sen
Balarka Sen
as $A$ is sent to the boundary circle by the snake to the boundary, right?
 
Exterior
Exterior
2:02 PM
@columbus8myhw google "connecting morphism"
yep
 
Balarka Sen
Balarka Sen
right, makes sense.
 
Exterior
Exterior
I'm thinking about this in terms of $Z_n(X,A)/B_n(X,A)$
 
Balarka Sen
Balarka Sen
actually i was thinking about doing it by the long exact sequence of the triple $(X, B, S^1)$.
 
columbus8myhw
columbus8myhw
Oh, it's something in category theory.
I don't know any of that stuff...
 
Balarka Sen
Balarka Sen
no it's not @columbus8myhw. it's about homology.
 
Exterior
Exterior
2:04 PM
basically, the point is that any relative cycle which includes more than just $B$ can be decomposed into $B$ and stuff in $A$
 
Balarka Sen
Balarka Sen
yeah.
 
Exterior
Exterior
and the stuff in $A$ does not affect the equivalence class in $Z_n(X,A)/B_n(X,A)$
I'm just not sure how to formalize this trivial-looking statement
 
Balarka Sen
Balarka Sen
well, it's the definition of relative cycles.
 
Exterior
Exterior
the definition just tells me this thing is a relative cycle
I want it to be equivalent to $B$
 
Balarka Sen
Balarka Sen
well, maybe look at nontrivial cycles in $A$.
 
columbus8myhw
columbus8myhw
2:06 PM
Wait, are homology groups like when you have a disk with two holes in it, and you get a group from the paths (loops) that can be created?
Or with a torus, and you get another group from the paths you can have.
Or am I thinking of something else?
 
Exterior
Exterior
those are fundamental groups (homotopy groups)
 
Balarka Sen
Balarka Sen
^
 
columbus8myhw
columbus8myhw
These words are too confusing.
 
Balarka Sen
Balarka Sen
big words are always confusing, until you study it.
=)
 
Exterior
Exterior
then even little words become confusing
:D
 
Balarka Sen
Balarka Sen
2:08 PM
for example, i find topological quantum conformal field theory confusing, because i don't know what the hell it is about
 
columbus8myhw
columbus8myhw
OK
 
teadawg1337
teadawg1337
@Balarka $\frac{1}{1-\frac{1}{p^2}}=\sum_{k\ge{0}}\frac{1}{p^{2k}}\implies \prod_{p\in{P}}\frac{1}{1-\frac{1}{p^2}}=\prod_{p\in{P}}\sum_{k=0}^{\infty}\frac‌​{1}{p^{2k}}$
 
Balarka Sen
Balarka Sen
eh
how's that supposed to help?
 
teadawg1337
teadawg1337
Yuck
 
Exterior
Exterior
@BalarkaSen I'll be in the AT room
 
Balarka Sen
Balarka Sen
2:11 PM
@teadawg1337 why don't you assume there are finitely many primes?
and see what that gives you?
 
teadawg1337
teadawg1337
Ah, good point
 
Balarka Sen
Balarka Sen
both works.
but yeah, the latter is easier.
 
ys wong
ys wong
Hey guys
 
columbus8myhw
columbus8myhw
Hello!
 
ys wong
ys wong
I wanted to study Quantum Mechanics
 
columbus8myhw
columbus8myhw
2:16 PM
Someone's talking about complicated stuffs and someone else is talking about less complicated stuffs.
 
ys wong
ys wong
Do u know which areas of maths should i brush up on
 
columbus8myhw
columbus8myhw
I think I'll put you in the "complicated stuffs" category.
 
ys wong
ys wong
This is what i will be studying for quantum
 
columbus8myhw
columbus8myhw
I think, if you can't do one of the problems that they give you at the start, that's what you should brush up on.
 
ys wong
ys wong
How much maths do i need for that
 
teadawg1337
teadawg1337
2:25 PM
@Balarka Derp, this is a simple proof by contradiction. $\prod_{P}\frac{1}{1-\frac{1}{p^s}}:=\sum_{n=1}^{\infty}\frac{1}{n^s}$
Setting $s=1$ and assuming a finite number of primes implies that the left side would converge, and the right side would also converge. However, the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, which implies infinitely many primes.
 
Balarka Sen
Balarka Sen
Yep.
You can do this for $s = 2$ by noting that the left side is product of finite number of rationals, i.e., rational, while the right side is $\pi^2/6$, which irrational.
Contradiction.
 
teadawg1337
teadawg1337
Fascinating stuff
 
Balarka Sen
Balarka Sen
The point is that as you close towards the singularity $s = 1$, you see $\zeta$ revealing a lot of information about primes.
This raises the question of whether closing towards other singularities, say, $\Re[s] = 1$ would reveal further information about primes.
It indeed does, but the calculations are a wee bit tedious than what you have just done. In fact, one derives the prime number theorem from inspecting $\zeta$ around $\Re[s] = 1$
 
teadawg1337
teadawg1337
I'm all too familiar with tedious calculations :P
 
Balarka Sen
Balarka Sen
And, of course, "jumping" to the region $\Re[s] < 1$ where $\zeta$, in the usual sense, is all nonsensical, is all about Riemann Hypothesis.
@teadawg1337 You can have a look, but it assumes that the reader is familiar with a bit of complex analysis, so you might study that first.
 
Alyosha
Alyosha
2:42 PM
How does one prove that, in a number field $L$, if $p$ is a prime, $\langle p \rangle$ ramifies iff $p | D_L$?
Maybe there needs to be assumptions that $p\ne 2$ also, I don'r remember.
 
Chris's sis
Chris's sis
3:38 PM
@teadawg1337 Compute this one without pen and paper $$\int_{-\infty}^{\infty} \frac{e^{a x}}{1-e^x} \ dx, \space 0<a<1$$ (it's not a joke, it's for real)
Greetings (btw)
 
teadawg1337
teadawg1337
@Chris'ssis Hello there :D
 
Chris's sis
Chris's sis
:D
 
Semiclassical
Semiclassical
i guess my instinct would be to drop the odd part of that integral, and see how the (even) integrand looks as an integral from 0 to infinity
 
Chris's sis
Chris's sis
@Semiclassical Each one separately diverges. In his book Inside Interesting Integrals, Paul J. Nahin do it by complex analysis.
 
Semiclassical
Semiclassical
hrm. so to the extent that the odd part goes to zero, it's only in a cauchy sense
yeah, i was about to say that that's probably the right way to think about it
figure out how to make a useful closed contour, work out the residues of the poles at $x=i \pi n$ for positive integer $n$ and so convert it to a sum
 
Chris's sis
Chris's sis
3:50 PM
@Semiclassical rectangle contour works fine with some tiny indents.
 
Semiclassical
Semiclassical
right. in which case i don't need all those poles
so, yeah. the integral from $x=\infty+i \pi$ to $x=-\infty+i \pi$ should do the trick, and it gives the same value as the original integral up to a prefactor
so adding it to the original integral gives a closed contour around a simple pole, and you're basically done
 
Chris's sis
Chris's sis
@Semiclassical You need no pole actually. You avoid them by your contour.
(just apply the Cauchy's first formula)
 
Semiclassical
Semiclassical
i see your point. the poles are at $x=0, \pm 2\pi i,$ and so forth
 
Chris's sis
Chris's sis
Yeah.
 
ɧɿρρԹʅȝՇԵՐՎԾՌ
ɧɿρρԹʅȝՇԵՐՎԾՌ
4:20 PM
Who's the ice cream guy ?
 
infinitesimal
infinitesimal
4:51 PM
He's looking for a cone :-)
 
Chris's sis
Chris's sis
5:25 PM
@ɧɿρρԹʅȝՇԵՐՎԾՌ compute the integral above without pen and paper ... (is it possible?)
Ah, he's out ...
 
JSG
JSG
6:07 PM
Is it true that X_t is normally distributed if X is a martingale?
 
A Beautiful Mind
A Beautiful Mind
Hi @ted.
 
infinitesimal
infinitesimal
6:24 PM
How are you today @abe?
 
A Beautiful Mind
A Beautiful Mind
Right now, bad.
 
infinitesimal
infinitesimal
Oh? Why.
 
A Beautiful Mind
A Beautiful Mind
The usual things bothering me, hard to describe.
 
infinitesimal
infinitesimal
Try.
:-)
 
A Beautiful Mind
A Beautiful Mind
Try to what?
 
infinitesimal
infinitesimal
6:27 PM
To describe what's bothering you.
 
A Beautiful Mind
A Beautiful Mind
I will try to get better, but I won't try to describe it here.
 
infinitesimal
infinitesimal
I see.
 
A Beautiful Mind
A Beautiful Mind
6:52 PM
Rehi @ted.
 
teadawg1337
teadawg1337
Rehi?
 
A Beautiful Mind
A Beautiful Mind
He seems to be in this room but is quiet.
 
teadawg1337
teadawg1337
Heya @Ted!
 
Huy
Huy
Sup.
 
teadawg1337
teadawg1337
@ABeautifulMind Perhaps he is busy, since it's around time for midterms at many institutions (I would assume)
Hello @Huy
 
Huy
Huy
7:03 PM
Hi teadawg.
Midterms.
I'm glad I don't have any midterms.
 
Swapnil Tripathi
Swapnil Tripathi
7:16 PM
I am having problems understanding the topology on \mathbb{R}^\mathbb{N} any help will be appreciated..
@TedShifrin
 
zed111
zed111
Hi all. Can you help me on this problem on number theory: For a fixed integer n>1, show that all the solvable quadratic congruences x^2 ≡ a (mod n) with gcd(a, n) = 1 have the same number of solutions.
 
robjohn
robjohn
@zed111 $\phi(n)/2$?
 
zed111
zed111
@robjohn Do you know the proof?
Oh got it. It is using the primitive roots, right?
The powers of primitive roots form Reduced Residue System mod n and furthur using Euler criterion only half of them will satisfy
 
robjohn
robjohn
7:32 PM
Suppose we have $b^2=a\pmod{n}$, then $x^2-a=(x-b)(x+b)$, so we should have one solution for each factor of $n$ (if $n$ is prime, we have two solutions).
 
Balarka Sen
Balarka Sen
@SwapnilTripathi What's your topology?
 
Swapnil Tripathi
Swapnil Tripathi
I recall R^N={f:N\to R : ||f||<\infty}
 
David Wheeler
David Wheeler
@robjohn ODD primes
 
Swapnil Tripathi
Swapnil Tripathi
@BalarkaSen
 
infinitesimal
infinitesimal
Hi pal @part
 
Balarka Sen
Balarka Sen
7:37 PM
yes, but what is the topology?
 
robjohn
robjohn
@DavidWheeler okay... yeah.
 
Mike Miller
Mike Miller
He's got a norm, @Balarka, so the one induced by the norm.
 
Swapnil Tripathi
Swapnil Tripathi
Product topology
 
Mike Miller
Mike Miller
Or not.
 
Balarka Sen
Balarka Sen
@SwapnilTripathi huh?
:P
 
David Wheeler
David Wheeler
7:37 PM
I know, 2 should be called something special cos it just so weird.
 
Swapnil Tripathi
Swapnil Tripathi
I guess? I'm so lost.
 
Balarka Sen
Balarka Sen
Topology doesn't come canonically, @SwapnilTripathi. You pick a topology.
In this case, I don't see what your topology is.
 
Swapnil Tripathi
Swapnil Tripathi
Our teacher defined that in the class and never talked about it later. I couldn't find any online resource to it @BalarkaSen
 
Mike Miller
Mike Miller
and what did your teacher define it as...?
 
David Wheeler
David Wheeler
@SwapnilTripathi You have a set of functions-$f:\Bbb N \to \Bbb R$, you are probably using some specific topologies on $\Bbb N$ and $\Bbb R$, it would be nice to know which ones.
 
Balarka Sen
Balarka Sen
7:40 PM
Specific topologies of N and R do not induce a topology on R^N.
 
David Wheeler
David Wheeler
@BalarkaSen No, but it's a start
 
Swapnil Tripathi
Swapnil Tripathi
@MikeMiller: he defined R^N to be the set of functions from naturals to reals and finite norm.
 
David Wheeler
David Wheeler
For example, we might be using the relative topology on $\Bbb N \times \Bbb R$, given the specific ones mentioned.
 
Swapnil Tripathi
Swapnil Tripathi
Never talked about topology.
 
Balarka Sen
Balarka Sen
What's the norm you are talking about?
Probably he just means the topology induced by the norm.
 
Swapnil Tripathi
Swapnil Tripathi
7:41 PM
Euclidean.
 
Balarka Sen
Balarka Sen
?
How is $||f||$ defined?
 
Mike Miller
Mike Miller
one normally equips function spaces with the compact-open topology. when the domain is discrete, this agrees with the product topology, @Balarka.
@BalarkaSen I suppose he means $$\|f\| = \left(\sum_{n=0}^\infty |f(n)|^2\right)^{1/2}.$$
 
Swapnil Tripathi
Swapnil Tripathi
f(n_1,n_2,....)=(x_1,x_2,...)

Now \sum_{n\in N} x_n^2 <\infty
 
Balarka Sen
Balarka Sen
Yugh.
 
Mike Miller
Mike Miller
this is also known as $\ell_2$.
 
Balarka Sen
Balarka Sen
7:43 PM
OK, so this topology is just coming from the metric induced by the norm
What's not to understand, @SwapnilTripathi?
 
Mike Miller
Mike Miller
it's a hilbert space, @balarka. it's the least yugh thing you could get.
 
Swapnil Tripathi
Swapnil Tripathi
Ok, I've never heard that word (in class). Talking about hilbert space
 
Balarka Sen
Balarka Sen
@Mike Never studied functional analysis a lot, myself. It's in the back of Simmons, so I guess I could read about it a bit.
(Motivation coming from Grothendieck :P)
 
Mike Miller
Mike Miller
@SwapnilTripathi don't worry about it so much
 
Ted Shifrin
Ted Shifrin
Good night, @Mike.
 
Balarka Sen
Balarka Sen
7:46 PM
@Ted!!
 
Mike Miller
Mike Miller
morning ted.
 
Swapnil Tripathi
Swapnil Tripathi
Just before this question he was talking about R^n with product topology. More easy to visualize and work on, maybe. Suddenly he talks about R^N and I feel scared already
 
Balarka Sen
Balarka Sen
ps : i don't know how to show that degree of the map induced by the vector field is 0, @Mike
 
Swapnil Tripathi
Swapnil Tripathi
@BalarkaSen
 
Mike Miller
Mike Miller
that's a shame, @balarka
 
Ted Shifrin
Ted Shifrin
7:47 PM
Hi @Balarka, @teadawg, @Swapnil. Oh, Swapnil, your question is already being addressed. There are lots of topologies.
 
infinitesimal
infinitesimal
Hello Proffessor @ted
 
Balarka Sen
Balarka Sen
R^N has no relation with R^n whatsoever
 
Ted Shifrin
Ted Shifrin
hi skull
 
Swapnil Tripathi
Swapnil Tripathi
Hello sir @TedShifrin
 
Ted Shifrin
Ted Shifrin
I hate that notation, btw ... Please make the $N$ into $\Bbb N$.
 
Balarka Sen
Balarka Sen
7:47 PM
@MikeMiller =(
 
A Beautiful Mind
A Beautiful Mind
Rerehi @ted.
 
Ted Shifrin
Ted Shifrin
LOL, hi, Jasper
My office hour isn't too crowded. One student parametrizing a tilted circle.
 
Swapnil Tripathi
Swapnil Tripathi
What would be the open sets here? :/
 
Mike Miller
Mike Miller
You're chatting during office hours, @Ted?? Shame on you.
 
Balarka Sen
Balarka Sen
lol
 
Theorem
Theorem
7:49 PM
@SwapnilTripathi Where ?
 
Mike Miller
Mike Miller
@SwapnilTripathi A basis for the open sets of a normed space are $B(u,r) = \{v : \|u-v\| < r.\}$.
 
Balarka Sen
Balarka Sen
open balls in the induced metric @SwapnilTripathi
 
Ted Shifrin
Ted Shifrin
Well, @Mike, not when I have 10+ students.
 
Balarka Sen
Balarka Sen
*unions of
it should be pretty easy to enumerate.
 
infinitesimal
infinitesimal
What is the greatest number of students that has attended one of your office hours @ted?
 
Ted Shifrin
Ted Shifrin
7:51 PM
I have had to move to a classroom, skull ... close to 20.
It's because I can't teach :D
 
Balarka Sen
Balarka Sen
I will be following your multivariable calc lectures when I get to it, @Ted
 
Ted Shifrin
Ted Shifrin
you'll hate it, @Balarka :P
 
Balarka Sen
Balarka Sen
well, let's see ;)
 
Ted Shifrin
Ted Shifrin
some will be way too slow for you, but you can skip minutes ...
 
Swapnil Tripathi
Swapnil Tripathi
I'll be back in some time @BalarkaSen. Going to read a little on it (now that I know what it is called )so that I don't talk silly.
 
infinitesimal
infinitesimal
7:55 PM
Please reconsider doing a MOOC @ted you just have to combine and edit your already excellent YouTube videos?
 
A Beautiful Mind
A Beautiful Mind
@infinitesimal MOOC?
 
infinitesimal
infinitesimal
Massive Online Open Course
 
robjohn
robjohn
@zed111 My proof works for odd $n$, but the statement is false for $n=6$: there are two solutions for $x^2=1:\{1,5\}$, two solutions for $x^2=4:\{2,4\}$, but only one solution for $x^2=0:\{0\}$, and one solution for $x^2=3:\{3\}$.
 
Chris's sis
Chris's sis
@ABeautifulMind ^^^ (to be the best one is not necessarily my aim, but to be incomparable in the area of integrals, series and limit as Ramanujan is :-))
 
A Beautiful Mind
A Beautiful Mind
@Chris'ssis Hey thanks. I feel like shit now.
 
Ted Shifrin
Ted Shifrin
7:58 PM
Actually, skull, as it stands, I don't even have access to that youtube page.
 
infinitesimal
infinitesimal
:-O
 
Chris's sis
Chris's sis
@ABeautifulMind Watch the video.
 
infinitesimal
infinitesimal
But why?
 
Balarka Sen
Balarka Sen
wait a tick @Mike.
 
A Beautiful Mind
A Beautiful Mind
@Chris'ssis I am so afraid I may never get well, so afraid...
 
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