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20:00
Even Milnor makes the obligatory pun. "This approach to the problem runs into manifold difficulties."
@TedShifrin I swear one of these days someone is going to make a graphing calculator named manifold destiny.
Graphing calculators aren't too good with manifolds.
wait
nvmd
for some reason your pun confused me
hard to graph something more than 2 dimensional
and I thought "manifest destiny" was "manifold destiny"
so the pun was totally lost
XD
20:02
@TedShifrin i tried a few different things i said $ (1+\frac {1}{n})^n=c $ then tried to ln both sides and take the limit of n to infinity to try and show that ln(c)=1 so $c=e^1 $ but it just didnt feel like it i wasnt breaking some logarithmic law when i was putzing around with it. i think it might be easier to show $ (1+\frac {x}{n})^n = e^x $ at least then i could use the MVT and FMToC any hints how to approach this better?
What's $\lim\limits_{h\to 0}\dfrac{\ln(1+h)-\ln(1)}h$, @Faust?
If I believe Akiva's intuition, I should look for framed torii in S^4 which are not framed cobordant to sphere.
@Faust: Then set $1/n = h$ and see this is the same thing.
@TedShifrin is the arf invariant used when making subwoofers?
Hey, why is there math.stachexchange.com and mathoverflow.net? What are the key differences?
20:05
Latter is mathoverflow, the former is mathunderflow
@red_trumpet mathoverflow is intended to be an eventual replacement. snigger
in all seriousness
MO is for research or reeeeeaaally complicated things
MSE is for people wanting help with this, that, and their mother.
@BalarkaSen Sure, that's what you're doing
my problem is oddly what the $ \lim_{h\to 0}\frac {\ln (1+h)} {h}$ is it looks undefined to me
@Faust: It should look a lot like the definition of the derivative.
I am... underflowing?
20:06
it does
@BalarkaSen no
So what's $\dfrac d{dx}\ln(x)$?
underflow is completely irrelevant
hi guys. I want to a paragraph that basically introduces my passion and understanding in mathematics.
Math overflow is a reference to stack overflow
20:07
$\frac {1}{x}$
any suggestions?
and stack overflow is a computer science bug made into a cute name
And at $x=1$ that is ... ?
@BalarkaSen trying to find tori etc
@Typhon ok, thank you. I actually like the design of MSE better, would be sad to find it replaced...
20:09
@TedShifrin is is undefined
@red_trumpet except MO is a paid service
or used to be back in the day
people used to have to pay to create communities on here
@MikeMiller right, i was making a stupid joke
this site may one day become....
Imagine that.
Pay per view
hey, question for y'all: if $n \equiv 0$ (mod $3$) and we are considering polynomials over $\mathbb{F}_3[x]$, is it true that $\text{gcd}\,(x^n-1, x^m+x^k+1) \mid x^{n/3}+x^{2n/3}+1$ for every choice of distinct $k,m \in \mathbb{Z}/n\mathbb{Z}$?
20:10
It would curtail all the homework-askers.
So, it's interesting to think about this. I don't know if it's visually apparent to me of such a framing on the torus (say)
@TedShifrin yeah but guess what, it is 10 cents per post here.
we'd all be bankrupt
jk
actually we'd all leave
D:
I wouldn't be. Answerers wouldn't have to pay.
@TedShifrin no 10 cents per chat post
and I mean 99 cents per every page load
I'm outta here.
20:12
@TedShifrin heheheh
glad that is unlikely
but fun to speculate
Well, in some ways, that's how life is if you want to read some papers the legal way.
If you university isn't subscribed to the journal it was published in, you have to pay.
@SteamyRoot Unless it is on arXiv, which it usually is these days
@SteamyRoot pay per google
True, and that's a lifesaver.
pay per email
20:16
But for papers that aren't on arxiv and you're not subscribed to, you pretty much only have the title and abstract to tell you what's inside.
Sometimes that's fine, but if you're looking for an obscure lemma or example...
@SteamyRoot pay per page; pay per reload; pay per operation in calculator
I keep coming up with more ideas to torment people with horrible real life drm.
@MikeMiller @TedShifrin Ok, turns out the framed cobordism group of surfaces in $\Bbb R^N$ is generated by the torus with it's Lie group framing. This was proved by Pontryagin, using the Arf invariant.
@AkivaWeinberger I guess that means generic preimage of your map is indeed the torus.
I have no idea how to prove any of this and would love a reference.
Google andy putman pontryagin thom
3
Q: Proof for elements of $\textbf{Z}[\sqrt{3}]$ regarding the existence of the norm.

TyphonSo for some context, I was in a proof writing class a couple months back. I really liked it and did quite well, but midway through the course we were doing things regarding the norm of these other kinds of integers (elements of $\textbf{Z}[\sqrt{3}]$). Basically things like the fact that there is...

Got it, thanks.
20:28
correct me if I am wrong, but would any ring created by appending solutions to polynomials also have the same norm property assuming it had unique prime factorization up to units?
@Typhon Not sure what you mean by "norm property"
read my question
it should be clear from context given what the question is asking
No, it is not at all clear what "norm property" you are asking about here.
Just parked a car for the first time in my life, scary experience, but went better than expected
20:32
You need to drive like Arturo Benedetto Giovanni Guiseppe Pietro Archangelo Alfredo Cortaffoli da Milano [EDIT: minor typos]
who even cares about parking?
Parking is like the brakes, superfluous?
exactly
Does anyone have an intuitive explanation for the central limit theorem? It feels way too much like dark magic to me
could anyone help me write a short sentence that captures something cool about mathematics or why it is interesting?
@Alessandro put a bunch of stuff in a pot and it diffuses out and loses all edges
20:40
@Dis-integrating "Math is cool"
I hate writing but if I must.
i feel like you're going to write something here
or was that it?
either way, thanks
I'm actually dying here though
I want to apply to cambridge
@s.harp eh
Not sure how serious that was, but I didn't get it
@s.harp lol
Hello world
20:47
Man I have not been on in years
@TedShifrin You still using MO?
wOW
I guess I don't really come on math.se anymore
i don't recall you
from here
Bro I was active in like 2011-2012
@BalarkaSen e.g. this was me: math.stackexchange.com/a/178381/165083
@fpqc yer a newb
@BalarkaSen I deleted my account
Then made a new one
20:51
welcome back, even though i know you from my past life only it seems
dont lie....
I am benjalim
no you are not
14
A: for a $3 \times 3$ matrix A ,value of $ A^{50} $ is

user1551You should learn BenjaLim's answer, which provides a general method for dealing with this kind of problems. However, here is a simple answer just for fun. Note that $$ A^2= \begin{pmatrix} 1&0&0\\ 1&1&0\\ 1&0&1\end{pmatrix} =I+\underbrace{\begin{pmatrix} 0&0&0\\ 1&0&0\\ 1&0&0\end{pmatrix}}_{L}...

and I'm Donald Trump... -_-
20:52
@BalarkaSen They mention benjalim.
liar
Ah, Ben Lim.
I have seen you on MO then.
Yes duh....
I don't know why my chat name is still fpqc
it's weird
maybe because it's still linked to your deleted MSE account
you can probably associate it to your MO account
20:53
yes
you're not ben lim
stop lying
In any case I feel like math.se is dominated by lots of stupid questions nowadays
homework-like stuff. i agree
It's getting boring.
Also the questions I usually have are more appropriate for MO
MO is too hard for me
20:55
Sure. If you're a grad student like me it's ok.
@TedShifrin Long time no see! @robjohn
@TedShifrin From memory your advisor was chern yes?
@BalarkaSen are you an undergad?
more or less pretty much almost.
@TedShifrin i think i maybe got it lets assume the limit could converge to some constant L so $L=\lim_{n \to \infty} (1+ \frac {1}{n})^n$ next ln both sides to yield $\ln (L)=lim_{n \to \infty} \space n\space \ln (1+ \frac {1}{n})$ if we let $ h =\frac {1}{n} $ then we have $\ln (L)=lim_{h \to 0} \frac{1}{h} \ln (1+h)$
or $\ln (L)=lim_{h \to 0} \frac{\ln (1+h)}{h}$ now the FToC should let $f^{'}(c) = lim_{h \to 0} \frac{\ln (1+h)}{h}$ now we know that the original function was $f(x)=\ln(1+x)$ so $f^{'}(c) = \frac {1}{1+c}= \ln (L) $ so $L=e^{\frac{1}{1+c}} $ now $ c \in \mathbb{R} $ but more than that $c \geq 0$ i think c is bounded above by $\frac {1}{n} $as $ n\to \infty $ so $L=e$ ?
(Also, yeah, Ted's advisor is Chern)
ayy @BalarkaSen which university?
the thing is im not sure i can ln a limit
20:59
your question is what is $\lim (1 + 1/n)^n$?
It's just e man
im trying to prove that it is e
Look, take the log of that
you have to prove the limit of (1+1/n)^n is 1.
l'hopital's rule then

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