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01:00 - 17:0017:00 - 00:00

01:00
@Semiclassical Oh I agree. I think you've raised some good points.
@TedShifrin This is the research that UWO offers for MSc and PhD Mathematics students... imgur.com/a/lhF5Y are those pretty broad topics, or is this a sign that UWO doesn't have a strong graduate math program?
Ok, I watched Eraserhead. God, what a movie.
Lmfao
01:25
I saw that 30+ years ago, Balarka. I barely remember anything.
That alone isn't enough to decide anything, Nate, and it's honestly not that relevant when you're going to be an undergraduate. Plenty of people go to good liberal arts colleges with no graduate program and turn into very good mathematicians.
I take it Balarka has given up on un-un-un-sleeping yet again.
@TedShifrin Yeah...
So would you say that time has erased the memory from your head?
duck
If I haven't slept, I'll always fall asleep during a movie.
You're presuming that I had memory to start with, @Semiclassic.
the whole point of scary shit like this is to keep me awake, really
01:30
I think this movie is my #1 from Lynch out of everything I watched so far
but I won't say "it's good" or "i like it" or "i don't like it" because none of those categories really apply for this one
Hmm, the movie was 40 years ago. So more than 30.
wow, really a long while ago.
you saw it when it came out?
I can't quite remember whether I saw it when it first came out or a few years later.
It might have been even 7 years or 8 years later.
Zee
Zee
02:03
@TedShifrin hello ted
hi chat
Zee
Zee
Chat is dead
:(
02:24
0
Q: Expansions of $\sum_{}^{}\frac{{\partial^{k}{f}}{ (\partial {z}^{k})(P)}}{k!}(z-P)^{k}$

ZophikelIn the text "Theory of Functions of a Complex Variable" I'm having trouble verifying the power series expansion for the theorem within $(1.)$ my initial attack can be seen in $(2.)$ $(0)$ $$(\frac{\partial}{\partial z})^{k}f(z)= \frac{k!}{2 \pi i} \oint_{|\zeta - p|=r}\frac{f(\zeta)}{|\zeta -z|...

Chat is alive again
 
1 hour later…
03:38
Hi
I have many long equations with so many parameters in it , in order to solve for a particular unknown it is little difficult to do by hand pages after pages, but yes it is ok though but still I need to verify it so
I know that Wolfram is supporting symbolic expressions so any other way how I can handle these ?
Zee
Zee
Blood sweat and tears
04:00
there are numerous computer algebra systems. look up CAS
$\log(1 - x + x^2) \approx -x$
?
thanks @Dair
is the above approximation true?
well... i mean it depends on what you mean by $\approx$
but...
if you look at series expansion at 0... it gives -x as the first term
so in some cases it would constitute as a good approximation near 0.
approx i meant
approx yes, but I am saying that the two functions are really only approximately equal near 0.
whoops...
they aren't that close the farther away you go.
nice one
yes
04:09
there are also other approximation methods besides for taylor expansions...
oh I see
04:35
any help on the convergence of ths series?
$(\cos(\frac{1}{n}))^{n^3}$
$n = 1 $ to $\infty$
looks like a limit not a series. also, use taylor series.
$\sum_{n=1}^{\infty}(\cos(\frac{1}{n}))^{n^3} = ?$
that is a series
$U_{n} = (\cos(\frac{1}{n}))^{n^{3}}$ of the series,

Taking $\log$ we get $\log(U_{n}) = (n^{3})\log(\cos{\frac{1}{n}}) = (n^3)\log(1 - \frac{1}{2n^{2}} + \frac{1}{24n^4} - ... ) \approx (n^3)(-\frac{1}{2n^2})$

So as $n \rightarrow \infty$, $U_{n} \rightarrow 0$ thus the series converges!
same response. use taylor series for cos(1/n)
04:39
is this correct?
yeah. basically ratio test.
ok so the series converges right.
but what is wrong in this ?
Using the fact that $a^{u} = e^{u.\log{a}}$

$e^{(\cos(\frac{1}{n}) - 1)^ {n^3}} = e^{e^{n^3.\log{(\cos{\frac{1}{n} - 1)}}}}$ next when $n \rightarrow \infty$ this goes to $e^{e^{(n^3)\log(-\frac{1}{2n^2}+\frac{1}{24n^4})-...}}\approx e^{e^{(n^3(-\frac{1}{2n^2})}} = e^{e^{(\frac{-n}{2})}}$,

now this tends to 1 (doesnot tend to $0$) as $n \rightarrow \infty$, hence the series diverges.
$\log(-1/2n^2)$ is not $\approx -1/2n^2$ as $n\to\infty$
cos(1/n)-1 is negative, by the way
still there is a power of $n^3$
which may or may not be odd!
what does $(\cos(1/n)-1)^3$ have to do with $\sum \cos^3(1/n)$ anyway, and why are you superfluously exponentiating $(\cos(1/n)-1)^3$?
04:53
ok lost in the echo
05:49
How do you enable the viewing of LaTeX in here?
see "LaTeX in chat" in room description on the starboard, upper right corner -->
Oh awesome! Thank you.
oh now i see
LaTeX why alternate capital and small letters?
any speciality in that!
SBM
SBM
06:25
Emphasis
I always dislike getting to messy solutions, since I know they're probably fine but there is a much better, simpler way. Then I get stuck
SBM
SBM
Um, I feel that if one is interested in a particular field she/he will do anything to learn interesting things about it
Don't worry if you get stuck
Eventually you'll get a lot better at solving, for practice and learning helps build one's abilities
@Studentmath
Yeah that's true, I guess.
Cheers
SBM
SBM
06:41
Pardon?
Thanks :)
Also, on the same matter - if anyone is interested in functional analysis, I posted a question where it's just what I was writing about:
https://math.stackexchange.com/questions/2298459/spectral-representation-integral-operator
SBM
SBM
Functional analysis? Seems interesting
Yeah, I like it
I have few gaps in my knowledge I have to fill on the run, but it's really interesting
SBM
SBM
I'm afraid I don't know about spectral representation yet.
That's actually nice. I often find these results all to be an extension of the ability to represent vectors, generally, as combination of orthonormal vectors.
SBM
SBM
06:59
Vectors are fun to play with at least from the little knowledge I have acquired from school
I wonder if I should just solve the s.d.e and plug it into the original integral equation.
@Studentmath Results from finite dimensional linear algebra tell you that diagonalizable operators are simultaneously diagonalizable iff they commute. So this basically works on Hilbert spaces too, and there's this principle that if you have an operator which commutes with taking second derivatives (or the Laplacian in higher dimensions) than a spectral decomp is provided by Fourier series (the eigenfunctions for the Laplacian)
Of course this interpretation really isn't rigorous if you're looking at $L^{2}$
Oh wow. That's cool.
Yeah, but perhaps I can find enough results in my textbooks to back up some more rigorous result.
Do you think, in this case of course, this is the more elegant way to solve it?
07:15
I've just glanced at the problem but the differential equations $f'' + \mu f = 0$ aren't hard to solve.
I can continue from there, my only issue is how do I know which of the eigenvectors I find are the one making the spectral representation?
Ah wait, I am stupid. Thanks!
SBM
SBM
08:07
Oh that ODE seems like easy @EricSilva
Though I hardly have any idea about Laplacian operator or spectral representation or eigen vectors
08:30
hey anyone up for computational discussion
like solving symbolic linear equations in computer
SBM
SBM
08:40
Ok seems interesting
08:58
Qu'il est tard ... @Ted
SBM
SBM
Pardon @Astyx
SBM
SBM
I couldn't understand what you just said.
No need to :)
SBM
SBM
Ok
Good afternoon bte
09:10
bte ?
09:20
betty?
hi @Astyx
09:41
Hi @SoumyoB
10:37
Guys, we know that for a function with period $2\pi$, it holds that
$$
c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)\exp(-int)\,dt.
$$
I would like to show that for a function with period $T$, it holds that
$$
c_n=\frac{1}{T}\int_{-T/2}^{T/2} f(t)\exp\left(\frac{-2\pi int}{T}\right)\,dt.
$$
I was considering the following substitution: $u(t)=\frac{2\pi}{T}t$. It seems to me that the only way for this to work is if we have the following:
$$
c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f\left(\frac{T}{2\pi}u\right)\exp(-inu)\,du=\frac{1}{T}\int_{-T/2}^{T/2} f(t)\exp\left(\frac{-2\pi int}{T}\right)\,dt.
@Semi ah right, well I'm only supposed to read chapter 1 and 2 from Griffiths for my final this week, which means I know nothing about "formal" quantum mechanics yet. The formalism happens in chapter 3.
$c_n$ is the $n$-th term in the Fourier series for the function you're considering, right?
yes that's correct
maybe I shouldn't be thinking in terms of integration by substitution?
Hey guys! needed some help in simplifying the homogenous gravitational field metric to the Rindler metric. Suppose we have the metric: (1+bk)^(2)dt^(2)-dk^(2)=ds^(2), under the condition that bk<<1, i.e., in the non-relativistic limit, the metric represents the homogeneous gravitational field. I am trying to change this via coordinate transforms to the rindler metric: ds^(2)=r^(2)dt^(2)-dr^(2).


Unable to get the transforms right.
11:27
Is there a biology room? i can't find it
SBM
SBM
@AbdullahUYU
I had to leave for lunch @sha
SBM
SBM
oh
hope you had a good lunch @AlessandroCodenotti
How did you define the Fourier series? Because what you want to prove is a consequence of $\left\{ \frac1{\sqrt{L}}\text{exp}\left(i\frac{2\pi n}{L}x\right)\right\}_{n\in\Bbb N}$ being a complete orthonormal family in $L^2([-\frac L2,+\frac L2])$ @sha
SBM
SBM
Fourier series?
11:36
sorry for all the pings, took me a while to get the LaTeX right
SBM
SBM
I guess I should search the Wiki
The imprecise and quick explanation is that Fourier series are a way to write a function $f$ as an infinite sum of trigonometric functions
@SBM thank you
SBM
SBM
So you could write any function as an infinite sum of trigonometric functions?
11:51
@Alessandro yea I've searched for a different source, where they actually start of with the more general case, so I'll just take that as the main "definition"/theorem
and hi @Astyx
Hi @Sha
I have another question now though about Fourier:
they're saying that the upper integrand is 1
I'm guessing it takes a "trick" to see this, but I'm clueless
I initially tried to show it by writing out the trigonometric identity
and I'm guessing that works fine, but apparently it should be possible to see it at once
upper integrand ?
yea I meant that one
sorry, I was slightly too lazy to type it out
Expand $e^{ix}$ as cos+isin
11:58
It's 1 when $m=n$
@Astyx but how do you see that quickly?
wait
of course
i'm so stupid
$e^{-x}$ is the inverse of $e^x$
why
i was looking at the second expression you know, when they had already written out the antiderivative
I forgot to look at the actual integrand
thanks
SBM
SBM
oh
that was interesting ...
why did they even write down the antiderivative:l
12:02
@SBM no, you usually do that with functions defined on an interval such that $\int_a^b|f|^2\text{d}x<\infty$
SBM
SBM
oh
12:21
oh?
SBM
SBM
Fourier series
SBM
SBM
:-\
12:39
Have you heard about Talbot @Danu? The notes are posted in the homotopy chatroom if you're interested.
SBM
SBM
homotopy?
SBM
SBM
am reading it
@user314159 So...?
"if" you're interested
12:47
Hmm
Trying to create a bijection from $\Bbb N$ to the set of (2-color) Turing machines $T$
I guess I can create injective functions both ways and use Schröder–Bernstein
@user314159 Don't get me wrong---I appreciate you telling me if you think it's something I might find interesting. But how did you come up with this? :P
just stumbled upon it
:-)
My favourite users lack again.
Ah, @Secret is around ... (lurking in the dark ...)
I didn't see @robjohn for a very long while. I wonder what is going on ...
He's really busy.
@user314159 OK user $\pi$.
13:00
:P
SBM
SBM
oh
SBM
SBM
so much terminology, I am such a noob. Just googled Schröder Bernstein
Can $2\times2=\{(0,0),(0,1),(1,0),(1,1)\}$ effectively be relabeled as just $4$?
SBM
SBM
What? ×_×
 
2 hours later…
user228700
14:39
@Abhishekstudent: Hi. U invited me here?
15:28
Does anyone know anything about basic knot theory?
@Semiclassical Could you take a look the question i asked:
https://math.stackexchange.com/questions/2293979/find-angle-alpha-in-this-triangle-question
15:46
Does anyone know why, if $L$ is a split link, then $\det(L) = 0$?
@Astyx hi
Hi
SBM
SBM
hi chat
@Astyx remember the question about $||f(x)-x|| \le 1$ ?
SBM
SBM
oh
16:10
I do
@Astyx @Daminark @Dodsy: Here's today's. "When the cartoonists sketched people in the comic shop, they ..." (4/1/5) OCAEWWRDDR
SBM
SBM
oh
@Astyx so there is a nice solution, which Ted showed me - defining $g(x) = x-f(x)$ , from the unit ball into itself, it has a fixed point
Yeah I thought of that afterwards
Hi @ted
are the jumbles always so punny? :P
16:13
drew a crowd?
:(
sorry:(
Yes, @Steamy.
People always ruin the fun.
Question: does $l_2$ is locally compact?
i'm not sure about the answer so i tried proving it.
if we take $x = (x_n)_{n \in \Bbb N} \in l_2$ , and a nbhd of $x $ , it contains a ball $B(x,r)$ , so i thought that if $\overline{B(x , r/2)}$ is compact then $l_2 $ is.
So , does $$\overline{B(x , r/2)}$$ compact?
Hi @Ted
16:15
Hi @Balarka, @Liad
@Lozansky
@TedShifrin Hi
Heya and sorry @TedShifrin
SBM
SBM
oh hey
$\int \frac{1}{f(x)} \, dx=?$
SBM
SBM
uh
16:19
@Liad are you asking if the space of square summable sequences is locally compact?
Hi @EricSilva.
yea
Hi @Ted
@Liad think about infinite sequences
of points in the unit ball
ok. you are trying to show me there are points with no compact nbhd ? @EricSilva
SBM
SBM
$$\int \frac{1}{f(x)} \mathrm{d } x = ?$$
What?
16:23
@Liad, that balls aren't compact, so it's not locally compact
closed balls
im not sure why this implies that. locally comapct =for each $ x\in X$ there is a compact nhbd that contains open nhbd of x.
I meant the closed unit ball
Every open neighborhood contains a closed ball, if there were a compact set that contains an open neighborhood of x, that ball would be compact.
but no balls are compact.
or if the space is Hausdorff, for each $x \in X$ and for each $U$ nhbd of $x$ there is a nhbd $V$ of x s.t $\overline V \subset U$ , and $\overline V$ is compact
16:25
@SBM Yes, what is that.
I stumbled upon this exercise recently : We know the generalised Bertrand Series, that is $$\sum{1\over n\ln (n) \ln(\ln(n)) \dots \ln^{\circ k}(n)^\alpha}$$ converges iff $\alpha \gt 1$ for all $k\in\Bbb N$. But what about $$\sum{1\over n\ln (n) \ln(\ln(n)) \dots \ln^{\circ f(n)}(n)^\alpha}$$ where $f(n)$ is the greatest integer such that $\ln^{\circ f(n)}(n) \ge 1$ ?
@EricSilva yea with that i agree, this is why i asked if $\overline{B(x,r/2)}$ is compact.
Zee
Zee
I hope algebra gets interesting. It's putting me to sleep now...
SBM
SBM
Is it even possible to integrate without knowing what f(x) is ?
right, and this is why I said to think about sequences :P
16:27
So, @Liad, Eric is trying to get you to focus on a particular case. Take $\overline{B(0,1)}$.
yea now i see what he meant.
in Hausdorff space compact is equivalent to sequentially compact right?
no, not true ... but you have a metric space
it's true in metric spaces
huh right.
ok , so i need to look for a sequence that does not have a convergent sub sequence
@Astyx Does the convergence fact about the first series come from comparing with the integral? (antiderivative of that thing is something like log(loglog...blah...log(x)) I think)
or something
16:31
Yup
Ok, cool
That's when $\alpha =1$
btw @Ted I've been reading Clelland and I quite like the exposition
@TedShifrin @EricSilva maybe the sequence $e_1,e_2,\dots$ ?
could be a good source of problems for people in the bootcamp who get interested like I did last year
16:32
When it's different than one you get a polynomial of degree $1-\alpha$ in what you just said
Yeah got it
Which is why it converges for $\alpha < 1$
@Liad, that's a good choice, why does it work?
Yup
SBM
SBM
oh
yeah, Eric, it's somewhere between undergrad and grad level ...
16:34
yeah if I had something like this right after I read the section in your notes on moving frames I would've devoured this book
well, I'm sure Jeanne will be glad to hear your praise :)
hi tern
I want to determine the flow of $\mathbf{A}=(x^2,2y,z)$ through the sphere $|r|=R$ by introducing the normal vector $\mathbf{n} = (x,y,z)/R$. This gives $$\iint_S \mathbf{A} \cdot \mathbf{n}dS = \dfrac{1}{R} \iint_S x^3+2y^2+z^2 dS = \dfrac{3}{R} \iint_S y^2 dS$$ due to symmetry. But how can I obtain $dS$ without changing coordinate system?
@EricSilva because if $x$ is the limit, then $||e_n -x|| \to 0$ , so $\sqrt((1 - x_n) \ ^ 2 + \sum_{i \ne n} x_i \ ^ 2 ) \to 0$ but this term is greater then $|x_n -1| $ so we have $x_n = 1$ for all $n$
So you don't want to use the divergence theorem, @Lozansky? Use symmetry some more!!
the sqrt should be until the $\to$
16:38
I don't see why $x_n=1$ for all $n$, @Liad.
Well @Liad by this argument you only have that $\displaystyle\lim_{n \to \infty} x_{n} = 1$, which is enough.
right @TedShifrin .
Please share the start chatjax link
it's right over there >>>>^^^^
16:41
Hi @Ted @Balarka @Liad
alright, so the unit ball is not compact , how do i conclude that each ball is not compact too? it seems right but im not sure how to justify it. maybe all closed balls are homeo' ?
Hi @AlessandroCodenotti
And @Eric and everyone else I'm forgetting
@TedShifrin thanks
Aren't you confusing euclidean norm and norm ? @Vrouvrou
Hi @Alessandro
16:42
Hi Alessandro
@Liad yeah, all closed balls are homeomorphic through a similarity, so if one isn't compact, none are.
(well, plus a translation, Eric)
yes yes you're right of course, I usually include rigid motions as similarities
Hi Amin
greetings, Demonark ... the jumble is above (but someone blurted out the answer, so ignore that)
16:45
great ,thanks @EricSilva , @TedShifrin
@TedShifrin It can be reduced to $$\dfrac{6}{R} \iint_S^{*} y^2 dS$$ where $S^{*}$ is the half-sphere
Not helpful, @Lozansky.
Hmm okay
Use symmetry the same way you did at the beginning, but a bit more cleverly.
16:47
@Astyx i don't understand
Drew a $x\in Sym(ocwdr)$
Oh crowd
Does the fact that the closed balls are homeo' is easy to prove? i dont see a nice way
Write down the bijection explicitly.
@Liad in which space are you working?
$l_2$
16:50
He's working in little $\ell_{2}$
sniped
@EricSilva welcome to the club
@Daminark im u now
Double sniped
wowowow.
@TedShifrin I think symmetry gives $\dfrac{3}{R} \iint_S R^2/3 dS$
16:51
Right
Ah okay
So $4\pi R^3$
@TedShifrin $z \in B(x,r) $ where to send it in $B(y,t)$ ?
You tell me, Liad.
$\ell^p$ is $L^p(\Bbb N)$ with the counting measure, right?
ok ok i tried :P
16:52
yeah @Alessandro
related @Daminark, last year in the bootcamp someone tried to do away with all the limit exchanging theorems in the complex analysis book by applying Fubini's theorem :P totally not in the right spirit
so don't do that lol
yes, @Alessandro
oops
Oh I think I might have seen his blog post or something?
idk, i do not remember who it was
@Astyx ?
If you can prove $f,g \mapsto f(0)g(0) + \int_{0}^1 fg$ is a scalar product you're done @Vrouvrou
16:58
if i don't know scalar prudact i can't prove this ?
@Ted I discovered this week that Cartan is really hard to read
$\dfrac{(1 + ||z-x||)v}{(1 + ||z-x||)t}$ maybe? @TedShifrin
Zee
Zee
@TedShifrin do you know Roy smith?
Élie, Eric? Yeah.
Yes, @Zee. We were colleagues for 30+ years.
01:00 - 17:0017:00 - 00:00

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