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6:00 PM
@LeakyNun Hey... do you maybe have an idea?
 
What class is this for @Evinda
 
It's number theory @Faust
 
analytic number theory?
hmm never seen any NT like that before.
 
@brot It follows that $|a_n - a_{n-1}| > \epsilon$ so given any bound $M$, there is always some number $N(\epsilon)$ such that a_{N(\epsilon)} > M$. Can you see why?
 
@Faust It's about primality testing. Do you have an idea?
 
6:03 PM
Sorry I meant: $a_{N(\epsilon)} > M$. Can you see why?
 
@Evinda sorry no, im actually a bit surprised i can even read the question.
 
Ok no problem :) @Faust
 
heya @Faust ... haven't seen you in weeks!
hi @Evinda.
 
Hello @TedShifrin
 
@berrygreen Isn't for $a_n = -1^n$ also $|a_n-a_{n-1}| \ge 1$ but it's bounded
 
6:06 PM
@TedShifrin and for good reason i aced my geometry final +p
 
that's great, Faust.
Well, actually, I was gone for 9 days.
 
O.o having too much irl then?
 
If I take derivative of a curve in a tangent space, which space does it live?
 
I was back in GA, cooking Thanksgiving dinner and visiting old friends, @Faust.
 
Is it still in the tangent space?
 
6:08 PM
Yes, @quallenjäger.
 
oh sounds like good time +)
 
Officially, it's in the tangent space at the appropriate point, but we identify the tangent space of a vector space (at any point) with the vector space itself.
 
@TedShifrin How can I argue that? Because of the $R^n$ structure of the tangent space?
I see
 
Somehow i got roped into invigilating a linear algerbra final
 
@TedShifrin bonsoir
 
6:09 PM
invigilating?@Faust!!!!
bonsoir, @Vrouvrou
 
@TedShifrin How comes that you speak french.
 
? that a bad thing?
 
Because I studied it for many years and majored in it (many years ago) ... and spend time in France when I can.
 
S'il vous plait, j'ai une fonction $\phi:\mathbb{R}_+\to\mathbb{R}_+$
1) non décroissante, continue à droite
2)$\phi(t)=0$ si, et seulement si $t=0$
3) $\phi(t)\to+\infty$ lorsque $t\to+\infty$
4) $\phi(t)>0$ lorsque $t>0$
5) $\phi$ est impaire

On définit $\Tilde{\phi}(s)$ par
$
\Tilde{\phi}(s)=\sup\{t\in \mathbb{R}_+, \phi(t)\leq s\}
$

avez vous une idée sur comment on peut démontrer que $\Tilde{\phi}(s)=0$ si, et seulement si $s=0$
 
In terms of set differences, what i $(A-B)-C$ equal to?
 
6:11 PM
I was just curious :D
 
Hey everyone!
 
Yeh
 
@Vrouvrou: Est-ce possible que $\tilde \phi(0)>0$ (ou $<0$)?
Heya Demonark (and Balarka). Demonark, I just got an email from another UC student a few hours ago.
@user193319: That is in terms of set differences. What do you mean?
 
@TedShifrin je veux montrer que $\Tilde{\phi}(s)=0$ si et seulement si $s=0$
 
J'ai vu ça. Il faut tout de même commencer.
 
6:14 PM
Huh, what was the email on?
 
@TedShifrin I meant, what is $(A-B)-C$ equivalent to? Is there a different way of writing it?
 
Tu vois que $\tilde\phi(0)=0$?
 
ah vous voulez dire par l'absurde?@TedShifrin
 
Oui, pourquoi pas?
 
Or is it confidential? In any event, cool
 
6:15 PM
No, Demonark. Just someone thanking me for the videos. Maybe someone in your class from last year now? I dunno.
 
@TedShifrin Meme commencer? That would be me
 
smacks Balarka
 
@TedShifrin hi !
 
hi @Liad :)
 
Oh wait actually I think I know who it is. Someone in my house is doing analysis (not honors, that got even crazier this year), and it's using a textbook which is garbage
 
6:16 PM
Which textbook?
even crazier? is that possible? that might warrant the roll of 13.123589732... eyes.
 
i need to show that $\int_{\gamma_R} \dfrac{e \ ^ {iz } -1}{z}dz \to 0$ when $R \to \infty$. ($\gamma_R (t) = R e \ ^ {it}$ $t\in [0,\pi]$ ).
Can someone help ? :P
 
Ted, you grew an irrational number of eyes!
 
I think it's called "A Course in Advanced Calculus" by Boller and Sally, it's incomplete and just bad.
 
This makes it to the history
 
Well, you aren't sure it's irrational, but it's likely.
@Liad: Did you write it out in terms of the parametrization?
 
6:18 PM
yea
 
Without the $-1$ it's pretty standard.
 
$ i\int_0^{\pi} e \ ^ {i R e \ ^ {it}} -1dt $
 
Yo, Eric.
 
@Daminark HA got crazier? how
 
6:19 PM
Well, what's the magnitude on $\gamma_R$?
 
So Soug's doing grad analysis and honors at the same time
 
Hi @Ted @Eric @Balarka @Dami and everyone else
 
 
And he finished Rudin 1-7 at just the same time that he started functional analysis
 
hi demonic Alessandro
 
6:19 PM
So he was like you know what? I'm gonna do functional analysis in 207 as well
 
@TedShifrin im doing this exercise, the $\gamma_R $ should cancel when $R \to \infty$
 
Oh great, Demonark.
 
But he only did like, 2 weeks
 
ah ok
 
Hey @Alessandro
 
6:20 PM
So it was like, open mapping theorem, closed graph theory, BCT, all that
 
wow my year was so different
 
But he never did stuff like weak convergence. So he started multi again after that
 
This is a standard example, @Liad, but it's a case where the "easy" estimate on the semicircle doesn't work.
 
@TedShifrin si je suppose que $s=0$ alors $\tilde{\phi}(s)=\sup\{t\in \mathbb{R}_+, \phi(t)\leq0\}=\sup\{t\in\mathbb{R}_+, \phi(t)=0\}=\sup\{t\in \mathbb{R}_+, t=0\}=0$
 
But you know, you can't actually do multi in a week
 
6:20 PM
@EricSilva: It gets worser and worser.
 
What's up?
 
So they're continuing it with Silvestre next quarter since he only reached Lagrange multipliers
 
@TedShifrin why did you say that without the -1 it is easy ?
 
Yeah, Demonark, it took me almost three quarters to "do" it correctly.
I didn't say it was easy.
 
silvestre is great so that would be cool to see
 
6:21 PM
we get $|e \ ^ {iR e \ ^ {it}}|$ in the integarnd
 
And what is that?
 
$e \ ^ {Re(iRe \ ^ {it})}$
 
hello friends
 
Yeah I think it was better for them to do functional with Soug and then multi with Silvestre, when Soug tried to teach multi my year it was a complete train wreck and none of us actually know multi because of that
 
OK. @Vrouvrou. Et si $\tilde \phi(s)=0$?
 
6:22 PM
@TedShifrin le probléme est si je suppose that $\Tilde{\phi}(s)=0$ cela veut dire que $\sup\{t\in\mathbb{R}^+, \phi(t)\leq s\}=0$ qu'est ce que je peux en déduire ?
 
i feel like our department needs a normal multi class
like very few people around here know multi and it's sad
 
Eric, it's worse than sad. It's criminal.
 
He just kinda taught differentiation poorly and then was like "Aight I want you to teach yourselves integration draws a blob you just cut this up into boxes and then do sup-inf like usual. Sounds good? Next class I'm gonna do integration on curves/surfaces"
 
Précisément, Vrouvrou.
 
@TedShifrin $|e \ ^ z| = e \ ^ {Re(z)} $ doesn't it?
 
6:24 PM
Est-ce possible que $s<0$ or $s>0$?
yes, @Liad.
 
he didn't do awful my year
maybe edwards is actually an ok book
 
so i just did the same with what we had @TedShifrin
 
Which Edwards?
@Liad: Write it as $e^{iR(\cos t+i\sin t)}$?
 
uh idk it's like "advanced calculus of several variables" or something
 
Maybe, but he told us that he seriously hated the book he used your year, and generally didn't like the way the class went for some reason
 
6:25 PM
if I have two holomorphic functions agreeing on a convergent sequence, what hypotheses do I need for the functions to agree?
 
and it's basically multivariable analysis
schlag really likes edwards
 
There's a C.H. Edwards book in Dover.
 
@GFauxPas They have to be equal everywhere
Identity theorem
 
f(z_n) = g(z_n) for (z_n) a convergent sequence
 
He was a colleague of mine. It's a good book with only one minor error.
 
6:25 PM
Oh, wait, are you agreeing on the limit?
Or no?
 
@TedShifrin je pense que je ne peux pas supposer que $s<0$ car $\phi(t)\geq 0, \forall t\in \mathbb{R}_+$
 
the identity theorem is on a compact set though/
?
 
Yeah
 
i think that HA works better if you just already know multi
 
so $= e \ ^ {Re(iR(cos(t)+ isin(t) )} $ @TedShifrin
 
6:26 PM
err connected
 
I guess you're not requiting $f(\lim z_n) = g(\lim z_n)$
 
I'm asking is that necessary/sufficient
 
requiring
 
Which is what, Liad?
 
$= e \ ^ {-Rsin(t)}$ @TedShifrin ?
 
6:26 PM
Right.
 
@GFauxPas No, if two holomorphic functions agree on a set with a limit point, they agree everywhere
 
So we have to show that $\int_0^\pi e^{-R\sin t}\,dt \to 0$ as $R\to\infty$.
 
yes
 
No connectedness necessary
 
why can we use the identity theorem if $(z_n)$'s image isn't connected?
 
6:28 PM
If you know some Lebesgue integration this is automatic. But otherwise you have to do a little work. This is sort of a famous problem in a beginning complex variables class.
Eric: But very few "know" it the right way, if any.
 
@GFauxPas It's part of the theorem statement. You can try to prove it
 
Does someone want to give me a two sentence primer on Symplectic Geometry? :P
 
@Liad: So the hint is: Do you know a convenient upper/lower bound on $\sin t$? whichever gives the right thing?
 
t?
 
@TedShifrin I guess that's true, although I've noticed a lot of my physics inclined friends are like really good at multi, seems like a decent way to learn it
 
6:29 PM
@Balarka so if you remember the quasifibration stuff I was talking about yesterday, turns out that's all you need to get the LES of homotopy groups
 
they're only really good at certain things, certainly not good at careful multivariable using linear algebra, Eric.
 
How would I prove it? Consider the set $\{z|(f-g)(z)=0\}$ and consider its accumulation points?
 
They're good a integrals over spheres and cylinders, though, which most mathies aren't.
 
@Daminark That's really interesting
 
sure not good at being careful maybe lol
 
6:30 PM
@TedShifrin t is an upper bound
 
is that the right idea Balarka?
 
We want an upper bound on $e^{-R\sin t}$. So what do we want here, Liad?
 
lower :P
 
@GFauxPas Mhm.
 
Right :)
 
6:31 PM
well, let's see
 
Recall that $p:E\to B$ is a quasifibration if $p:(E,p^{-1}(b)) \to (B,b)$ is a weak equivalence
 
Call that set $Z(f-g)$
 
So now you take the LES of the pair $(E,p^{-1}(b))$
 
Tee hee .. now Balarka is having to juggle multiple conversations :P
 
@TedShifrin :(
 
6:32 PM
$Z(f-g)$ can't contain limit points unless $(f-g) \equiv 0$ on the region where $f-g$ is holomorphic
 
@TedShifrin we can take the triangle below sin(t) when $t \in [0,\pi]$
 
... am I done?
 
$\ldots \to \pi_q(p^{-1}(b)) \to \pi_q(E) \to \pi_q(E,p^{-1}(b)) \to \pi_{q-1}(p^{-1}(b))\to \ldots$
 
Demonark: So are you telling me you told this guy to watch the videos? He seemed appreciative, regardless.
What do you mean, Liad?
 
the stright line between $(0,0) $ and $(\pi/2 , 1)$
 
6:33 PM
Ah. Good. Do it.
 
and from $(\pi /2 , 1) $ to $(\pi,0)$
yea that works :P
 
@Daminark Ah I see what you mean OK
 
By symmetry, you can just analyze the integral on $[0,\pi/2]$.
 
yea
but we changed the problem
 
OK, you're done. :)
 
6:33 PM
@Ted well, first off it's a girl I have in mind, and she found your videos on her own, I just remember her talking about them being helpful once, and she mentioned that her professor used sections of your book because theirs is horrible
 
Oh, OK, Demonark. This was a guy who emailed me.
 
i should integrate $\int_{\gamma_R} \dfrac{e \ ^ {iz} -1}{z} dz$ :/
 
Well, I'd be happy if UC started using my book. :P
 
@Balarka yeah at that point you can just sub in $\pi_q(E,p^{-1}(b))$ with $\pi_q(B)$ because of the weak equivalence business
 
we seems to use a lot of bad books
 
6:34 PM
but if $f(z_n) = g(z_n)$ and $z_n$ has a limit point, then $(f-g)(z_n)$ has a limit point , but $(f-g)(z_n) \to 0$?
 
@TedShifrin how can i bound this integral from above?
 
so $f \equiv g$?
 
the $-1$ makes things complicated @TedShifrin
 
@Daminark Yup
 
@BalarkaSen is that a complete proof
 
6:35 PM
@Liad: Do we really need the $-1$ to do the actual problem at hand?
They just put that there to make the integrand continuous at $0$?
 
their hint is to integrate this function
 
What sort of problems does a symplectic geometer care about?
 
I don't like that hint. Go back to the original question.
 
symplectic geometric problems
 
@GFauxPas Thanks.
 
6:37 PM
np happy to help
 
@GFauxPas Thanks.
 
so you say to integrate $e \ ^ {iz}/z$ would be enough? @TedShifrin
 
@GFauxPas Why can't $Z(f-g)$ have a limit point if $f-g$ isn't $0$?
 
@Ted I do think they should start using your book, though the issue is they run linear algebra and analysis as independent courses, and also in analysis they do want to do stuff like metric spaces, a complete review of sequences/series, uniform convergence/Arzela-Ascoli, and so forth
 
Yeah, since we want the imaginary part on the real axis, anyhow, Liad.
 
6:38 PM
@Balarka so yeah now I'm trying to prove that a (Serre) fibration is a quasifibration
 
uhh
 
Good exercise
 
Hi chat
 
Yeah, at UCSD they are using it and doing just the linear algebra chapters this quarter, and then the rest the other two quarters. It defeats the whole purpose of integrating the material (pun intended), but it works, probably.
 
if it has a limit point $\ell$, consider $f-g-\ell$?
 
6:39 PM
There is no book that does everything perfectly for every school's organization.
heya Tobias.
 
wait, no, i take that back
 
@TobiasKildetoft hi
 
@TobiasKildetoft Hi there.
 
This is true. I think the right balance for what they're looking for is Pugh's book, which is basically Rudin but with better-written stuff for multi
And lol that was good
 
I don't know @AlessandroCodenotti
 
6:40 PM
It is in fact true that an holomorphic function has a discrete set of zeroes, but that needs to be proved
 
Pugh is good
 
(unless it is always zero of course)
 
that's the identity theorem, we're allowed to use that
but what if $(f-g)(z_n) \to \ell \ne 0$?
 
Aren't you trying to prove the identity theorem?
 
Demonark: Pugh surely is way better than Rudin ... but still assumes people know multivariable calculus and can compute things. My book tries to teach all the multivariable in a self-contained way. There's a big difference.
 
6:42 PM
no, i'm trying to prove that, if holomorphic functions $f$ and $g$ agree on a sequence with a limit point, theyre identically equal
 
Pugh is way heavier than rudin is the problem
 
@GFauxPas But then that is clear as the difference has a non-discrete set of zeroes
 
I have a problem: $f(x+1)=3-2x; f(\varphi(x))=6x-3; \varphi(x)=?$ Any comments?
 
heavier? in what sense?
 
it's physically bigger
 
6:43 PM
@Ted I was actually under the impression that Pugh was self-contained, like it spends a lot of time on multi
 
Uhm what's your statement of the identity theorem? Because I saw it as "two holomorphic functions agreeing on a set with an accumulation point must agree everywhere"
 
It's an analysis text, Demonark. He assumes people know multivariable calculus.
Yes, he definitely spends more time on multivariable analysis, as is appropriate.
"self-contained" is a totally ambiguous term.
 
if two holomorphic functions agree on an open connected set, they agree everywhere
 
rudin's multi is effing baaaaad
 
Eric: Rudin is way too compact. It's a terrible book for most students.
Yes, Rudin was the antipodal point of a geometer.
 
6:44 PM
i actually dont like rudin for a class, i like rudin for if you have a load of time to puzzle over what he's saying on your own
 
ZERO pictures.
 
thats what I'm using @AlessandroCodenotti
 
Not good for the average student.
 
ears perk up
 
smacks Demonark's ears
 
6:45 PM
No pictures? I should relook at it then... maybe I'll understand it this time... :P
 
that's how i read rudin in hs and i think i got a good amount out of it
i hate the other books of his that ive looked at
 
Hm, ok, nevermind that. Pick an holomorphic function $f$ and $z_0$ such that $f(z_0)=0$, show that either $z_0$ is isolated or $f$ is zero on a nbhd of $z_0$
 
yes we did that in class :)
that was fun
 
Pugh does better with Lebesgue too, Rudin like starts doing Lebesgue theory and just doesnt get anywhere
 
oh, I guess that's it
 
6:47 PM
I don't understand why that chapter is even there
 
I would never have taught out of his Real & Complex, Eric, but I found it useful for me to read in grad school.
 
@GFauxPas Then you're almost done
 
in Homotopy Theory, 7 mins ago, by bob
A fascinating blog post : https://noncommutativeanalysis.wordpress.com/2017/12/05/the-nightmare/
 
I used it a lot for extra reading when i was taking grad analysis and I do not like it
i also hate his functional book
 
It's useful to intermingle the two somewhat, once you're a grad student. For example, dominated convergence totally nails the problem that Liad was working on in one second.
 
6:49 PM
so i consider $(f-g)(z_0)$
with $z_0$ a limit point
 
a good understanding of Fubini + DCT gets you a very long way
 
So, pick your holomorphic $f$ and its zero set $Z(f)$, consider $A\subseteq Z(f)$ the set of accumulation points of $Z(f)$, can you show that it is both open and closed?
 
then $(f-g) \equiv 0$
oh
 
(Of course $f$ needs to be holomorphic on a connected domain, otherwise the identity theorem is false)
 
it's closed because every point in it is an accumulation point
 
6:51 PM
Right, that's the easy one. You already did the other one actually
 
it's open because each point in it ... is...
 
@GFauxPas think about this
 
it's complement is closed
 
Hi DogAteMy
 
wait, thats circular
 
6:55 PM
Hi
Circles are closed yes
:P
 
Breaktime at school?
 
To show it's open you want to show that each point in $A$ has a nbhd contained in $A$
 
Well $A$ is open and thus a neighborhood, so each point in $A$ is contained in a neighborhood (namely $A$)
:P
 
we're proving that it's open Akiva
 
6:56 PM
Yeah, and I showed $A$ open $\implies$ $A$ open
An invaluable contribution to the team
 
Mazltov.
 
wtg
 
Wait a second @Balarka yesterday we were talking about the homotopy fibers and I'm no longer sure of what we were saying
 
If $X$ is some space, is $\{\emptyset, X\}$ is $\sigma$-algebra on $X$?
 
because it's a preimage of $\{0\}$?
which is closed?
oh but it might not be all of the preimage
 
7:00 PM
@user193319 Yes
that one is the important thing here @GFauxPas
 
oh
then each point is isolated
each point is both isolated and an accumulation point
whcih can't be, so each point has a neighborhood around it
but it's closed
so it's either the entire region or the empty set
?
so $f$ is identically zero or its zeros aren't accumulation points
we went backwards?
i still have to prove the case $(f-g) \to \ell \ne 0$
 
Ok, wait, so an holomorphic function which has a set of zeroes with an accumulation point is $0$ everywhere
Now you know that $f$ and $g$ agree on a converging sequence, by continuity they must agree on the limit as well so...
 
7:36 PM
so f(z) = g(z), and (f-g)(z) = 0, with $z$ an accumulation point :)
Awesome
thanks A
 
0
Q: Asymptotic expansion of the Gaussian hypergeometric function.

JuliusLet $N$ be a positive integer, $p_N=N^{-\alpha}$ for some $\alpha\in [1,2)$, and fix $k=0,\dots,N$. I would like to find such $f_{k,\alpha}(\cdot)$ that $$ F_{k,\alpha}(N) = {}_2F_1\left(-\frac{N-k}{2},-\frac{N-k-1}{2};1+k;p_N\right) \sim A_{k,\alpha}\cdot f_{k,\alpha}(N) \quad \text{ as } N\to...

 
out of all the math courses I've taken, complex analysis has the most surprising theorems
 
Yep, there are quite a few "too good to be true" theorems
 
I don't know what a function does. But I know what it does on a closed contour? Good enough for the interior!
holomorphic*
 
if $f$ is analytic $u$ is harmonic i need to prove that $u$ composition with $f$ is harmonic, but, what is $\dfrac{\partial f}{\partial x}$ ? this does not make sense to me because $f $ is "two" variables function
@GFauxPas i think we both studying complex analysis :P
 
7:46 PM
:)
can I apply Rouche's theorem to an entire half-plane?
$|f|<|g|$ on $\operatorname{Re}(z) = 0$, for example
 
@Daminark What is it that you are unsure of?
 
"Category theory is an island of beauty in the already beautiful land of mathematics, and we are in love with it. We shall communicate the bliss we feel when we do it." - Daminark
5
 
Top 10 anime dialogues of all time
 

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