« first day (2974 days earlier)   

8:00 PM
@Ultradark the branch of math that deals with this kind of stuff most concretely is perturbation theory
 
A sequence which diverges?
 
I'm forgetting what the best way to set it up in the present case would be tho
 
Everything is inside $E = \ell^1$, @Oskar.
 
Ah! Mea culpa.
 
Go reread your question.
 
8:00 PM
Will do!
 
@Ultradark take a look at this bit of the wiki page on Bring radicals I guess: en.wikipedia.org/wiki/Bring_radical#Series_representation
specifically the reference to solving the equation by 'reversion of Taylor series'
 
Yeah, we have $X$ explicitly written.
It's sequences with every second element being zero.
 
So what's a point in the complement of $X$?
 
I really think my approach is easier, lol
 
If you plug in a=-1 into their series, you'll get an infinite series for the solution of x^5+x-1=0
...assuming said series converges, though
ah, no it won't
 
8:03 PM
Leaky, hush.
 
A sequence $(x_n)_{n\in\Bbb{N}}$ such that $x_{2n}\neq0$.
 
"The series converges for $ |a|<4/(5\cdot {\sqrt[ {4}]{5}})\approx 0.53499$"
 
Quantifiers please?
Quantifiers please.
 
and a=-1 has |a|=1 which is definitely too big
so their method won't help here
 
$\forall n\geq1$?
 
8:04 PM
How do I negate the sentence "It rains every day."
 
It rains some day.
 
Um ...
 
Haha! It doesn't rain some day.
Sorry!
 
OK. So, now, what's a point in the complement of $X$?
 
A sequence for which there exists an $n\in\Bbb{N}$ such that $x_{2n}\neq0$.
 
8:06 PM
Good.
Let's call that sequence $(w_n)$, with $w_{2s}\ne 0$.
 
Alright!
 
Can you give me an $\epsilon>0$ so that the $\epsilon$-neighborhood of $(w_n)$ in $\ell^1$ is disjoint from $X$?
 
@Semiclassical
 
sup duds
 
Who're you calling a dud, Eric?
 
8:10 PM
$2\ln(x)=\ln(1-x)$ equals $1-\phi$ where phi is the golden ratio. that's pretty neat lol!
 
Functional analysis calls and you answer? Hi @Eric
 
it's cute, yeah
 
Well, only changing one element in the sequence which doesn't make it so that $x_{2n}=0$ for all $n\in\Bbb{N}$ will keep it outside of $X$, but I feel unsure about how distances work here.
 
@TedShifrin don’t worry everyone but you
 
@Ultradark actually, should be $\phi-1$
 
8:10 PM
OK @Eric
 
@AlessandroCodenotti lol all the functional i once knew i forgot
 
Well, how does a norm define distance, @Oskar?
 
oh yeah
 
@Eric: Hippa meme. Paging @Hippa!
 
Hi @Akiva
 
8:11 PM
which, thanks to the weirdness of the golden ratio, is just $1/\phi$
 
Normally you take the difference
 
howdy DogAteMy smack
 
two drums and a cymbal fall on my head
Heyo
 
@EricSilva Same to be honest... I wish there was a functional analysis course to take during my master but I there isn't one as far as I know
 
That seems unlikely, @Alessandro.
There's plenty of analysis at Bonn, isn't there?
 
8:13 PM
It's all dysfunctional
flees from the drums and cymbal
 
@AkivaWeinberger shalom
 
Well, we have the $\|\cdot\|_1$ norm.
 
ponders putting DogAteMy in timeout for particularly poor humo(u)r
 
i will eventually pick up a book (probably brezis again tbh) and read through it before grad school
 
indeed, @Oskar
 
8:14 PM
@MikeMiller oh i completely missed this
 
@Eric: If you take some serious PDE courses, you'll need lots of functional analysis.
 
@TedShifrin Well they have a lot of PDE courses which also include functional analysis, but I would be more interested in something like a course following Conway's book than one doing PDEs (Also the modules handbook is super vague, I'll ask more infos when I'm there)
 
@LeakyNun Shalom. Ani berakévet leáko
 
@loch so are you?
 
@TedShifrin i used a looot of functional this summer but nothing that wasn’t kind of easy
 
8:15 PM
And you've forgotten all your geometry in the meantime. There's the Hippa meme.
 
i popped in, got a cookie which i didn't like, and left
 
Okay. So, what's the epsilon then?
The sequences just need to differ by one element.
 
@loch: sounds like a typical antisocial grad student.
What size $\epsilon$ will force the open ball NOT to intersect $X$? And prove it, @Oskar.
 
:(
 
@TedShifrin at least i’m taking a geometry class!!
 
8:17 PM
LOL, yes, Eric. Or else I'd stop talking to you :P
 
i’d stop talking to me too tbh
 
N..
@Semiclassical @Ultradark a couple of Newton method iterations should do it
 
@loch let's be proud to be antisocial
 
if you're going to do any root-solving and you can use Newton, it's probably the right way
 
8:20 PM
I feel all lost in sequence space.
 
Can someone explain me what's global analysis? All I know is that it's "analysis on manifolds", but what are some important or deep results in this field for example?
 
Write down the definition for $\|(w_n)-(z_n)\|_1<\epsilon$.
A big one is the Calabi conjecture (proved by Yau in the late 70s), @Alessandro. Google.
Lots of stuff that Eric probably can tell you that his adviser and colleagues have done with minimal surface theory in the past decades.
@Oskar: If you want to picture things a bit, draw a horizontal number line, and at each integer $n\ge 1$ draw a vertical0 line in the plane. You can plot your sequence by picking a point on each of those vertical lines. Now how do you measure distance between two such?
 
@loch haha thats ok
@loch i had a red oreo which was weird
 
i got the chocolate coated one

didn't realise it's mint chocolate..
 
@TedShifrin I would say the first one is the uniformization theorem - perhaps phrased in terms of Riemannian metrics
 
8:26 PM
Hey everyone!
 
@loch rip
 
Hi Demonark
 
is there anything good at 4:30
 
hi Demonark
 
i see bjorn runs a seminar then
 
8:26 PM
@TedShifrin Hmmm, that looks super hard, probably because I don't know half of the words in the wikipedia article!
 
i applied to grad school to work with him and they were like nawwwww
 
@MikeMiller number theory seminar
think today the talk is about some topological methods in number theory (at least i saw "steenrod operations" in the abstract)
so you might be interested
 
Every point $x_1,x_2,\dots$ in the sequence $(x_n)$ would have a distance to the points $y_1,y_2,\dots$ in $(y_n)$. The distance between the sequences as a whole seems a bit abstract to me right now.
 
"Artin-Tate Pairing on the Brauer group of a surface"
 
It's the sum of those (vertical, in my description above) distances, @Oskar.
 
8:29 PM
oh I see you're right
I have decided I am too lazy tho
 
Bah @MikeM. How often are you visiting such a den of activity?
 
I'm always a bit amazed when I remember exactly how fast Newton's method converges
 
Okay. I get it, @Ted.
 
I had my students prove that in homework, @Semiclassic.
 
"the number of correct digits doubles on each step? lulwut"
 
8:30 PM
So now finish, @Oskar.
 
@TedShifrin not very
 
Hi handsome fols
folks*
 
(assuming that it converges in the first place, of course. important caveat there)
 
The way I told you to draw it, you can actually "draw" what an $\epsilon$ neighborhood looks like, @Oskar.
 
but I decided I could get work done over the next hour and a half that is probably more useful than the talk
 
8:31 PM
OK, @MikeM. I'll accept that. So bye :)
 
FINE
 
LOL
<--- good at making enemies
 
@loch I have also been deceived by this so I know your pain
 
Demonark: Did someone feed you a cookie with pork in it? :)
 
hi @KasmirKhaan
 
8:34 PM
Every day in the math department here there's tea and cookies, one day they had those oreos and I was like, wow this is new. And then it was mint
 
oh yeah, hi @Kasmir
 
@Ted thankfully, no, at least to my knowledge
 
@LeakyNun Leaky :D
@TedShifrin Ted :D
 
Like that? I haven't drawn any $\epsilon$ yet.
 
Cool, @Oskar, so how would you describe the $\epsilon$ ball around $(x_n)$ pictorially?
 
8:35 PM
one of the profs i knew used some form of netwon's method, something like quasi linear approximation or something
 
What do you have to draw on each vertical line?
 
for his research
 
A line that surrounds every $x_n$ at a specific distance, $\epsilon$.
 
nope ...
you're going to draw vertical intervals around each $x_j$ so that ... what ... ?
 
what is a non possible affine variety
 
8:38 PM
So that we don't have that $x_{2n}=0$ for all $n\in\Bbb{N}$. Equivalently, that there's an $n\in\Bbb{N}$ such that $x_{2n}\neq0$.
 
No, no, @Oskar. Not yet. How does the $\epsilon$ come in?
 
It's the sum of all distances between the points of the sequences.
$\sum_{k=1}^\infty|x_k-y_k|$
 
So if our intervals are symmetric around $x_j$, half the sum of the lengths of all the intervals should be less than $\epsilon$.
OK, NOW, choose an $\epsilon$ that'll guarantee that the $\epsilon$-ball around our $(w_n)$ doesn't intersect $X$.
 
That $\epsilon$ must be smaller than the difference from all $x_{2n}\neq0$. Otherwise, we could find a sequence which has $x_{2n}=0$ for all $n\in\Bbb{N}$ and thus be inside $X$.
 
How was our original $(w_n)$ specified?
 
8:44 PM
That there was one $n\in\Bbb{N}$ such that $x_{2n}\neq0$.
 
OK, I think I named it $w_{2s}\ne 0$ or something.
Can you use that to decide on $\epsilon$?
 
@TedShifrin I think a similar idea is used to prove the countable additivity of Lebesgue measure or something
 
Thus, $0<\epsilon<|x_j|$ is required if $x_j$ is the one.
 
(at least one)
 
Oh, don't confuse with a different meaning for $\epsilon$, @Oskar.
 
8:45 PM
Huh?
 
Think about our picture, and put $w_{2s}$ in the picture.
 
Yeah!
 
hi all
 
$\epsilon$ has to be smaller than the distance that can get $w_{2s}$ down to zero.
 
if something is not a possible affine variety is that bad
do we like affine varieties?
 
8:47 PM
We like all math.
 
Right, choosing $0<\epsilon\le |w_{2s}|$ does it, @Oskar, because even if we use up all our $\sum |w_n-y_n|$ in that one slot, we don't have enough room to go down to $0$.
 
"If a man can't run a marathon, does that make him bad?"
 
come on guys
 
It depends on context.
 
Okay. So what do we have now then?
 
8:48 PM
well what does affine even mean
 
You've proved (after a few hours) that $X$ is closed.
 
Maybe start with that.
 
I think I know what a variety is
 
What do you mean by "after a few hours"?
 
@TedShifrin What's the question?
 
8:48 PM
How long have we been doing this, @Oskar?
 
A while. Why?
 
Oskar posted it a while ago, DogAteMy. Scroll up to a big pasted-in display.
 
Do you imply that I'm stupid?
 
That's why I said a few hours. :)
LOL, no, I just implied it took a while.
 
Oh! Okay. :)
 
8:50 PM
The idea for $Y$ is analogous.
 
Right.
 
I have some gaps that cause me some problems. Like, I didn't know how the norm in sequence spaces works.
 
lol, I'm sitting in the student union here
 
I feel like that happens to me every now and then.
 
8:50 PM
and someone is playing Banjo-Kazooie music on the piano
 
You definitely have to play with concrete examples to understand definitions, @Oskar. I give that advice repeatedly.
 
Is $x_n$ like $(0,\dots,0,1,0,\dots)$?
 
Where's your $1$, DogAteMy?
 
Indeed. I hadn't seen the definition of the norm in sequence spaces. It wasn't covered in the chapter by Brezies.
 
you have a $1$ on an even spot
 
8:51 PM
(a,0,b,0,...)
 
Actually, all the even slots are $0$.
 
Oh, I see
 
You can have any summable sequence in the odd slots.
 
@Semiclassical I have a small E&M question:>
 
8:52 PM
just so long as it's not so small as to be quantum.
 
Nah, not yet
 
Set of all things where the even slots are zero, got it. And Y is the set of all multiples of $(2^{-2^n})_n$ or something
 
Maybe take it in hbar? (as to not interrupt the ongoing lesson)
 
No, no, DogAteMy.
In $Y$ each odd term determines the next term, but they're just linked in pairs.
 
8:54 PM
(a,a/2,b,b/4,c,c/8,...)
 
Oh, I misread.
 
Right.
 
Seeing that $X+Y$ is dense is rather sneaky, but cool.
 
In both cases, the odd terms of the sequence determine the entire sequence
 
8:54 PM
If something hasn't popped up before, and Wikipedia doesn't have a good article about it, then it's kind of hard to find a comprehensive source for the material.
Oh, I like cool.
 
There are lots of other books to look at, @Oskar. My library is depleted so I can't recommend too many specifically. Simmons has a nice book on analysis and topology that's quite readable.
 
And $X$ is closed because, for a point in the complement, you can take the maximum even coordinate as your $\epsilon$ and draw an $\epsilon$-ball
 
Alright!
 
and $Y$ is similar, you take the even coordinate that's furthest away from what it should be
 
No, DogAteMy, that won't necessarily work.
Oh, yeah, it will. Never mind.
We just picked some nonzero one.
 
8:58 PM
Ah, that's a better way, yeah
 
You guys can discuss $X+Y$ and why it should be dense. I'm out of here for now.
 
Okay. Thanks for the help, Ted. Talk to you later!
 
Anything that's eventually zero is in $X+Y$ I think
And that's dense, isn't it?
 
Oh, and that's all of $E$ by how $\ell$ works.
 
Not everything in $\ell^1$ is eventually zero
 
9:01 PM
Wasn't that what Ted said?
 
I thought $\ell^1$ was the set of sequences of things whose absolute values converge
Absolutely convergent sequences
Like $(1,\frac12,\frac1{2^2},\dots)$
 
@Daminark I like the gimmicky oreos a lot
I'm a sucker for a dumb gimmick
 
There is a cereal of them
Oreo O's
This is real
 
Haha, nice
 
I'm confused. How are you going to approximate (0,1,0,2,0,0,0,0,0,...)?
 
9:05 PM
Got to love free food
 
That if $\|x\|_1=\sum_{k=1}^\infty|x_k|<\infty$ then we have that $\lim_n (s_n-s_{n+1})=0$, and $\lim_n x_n=0$.
 
$(-1,0,-4,0,0,0,\dots)+(1,1,4,2,0,0,\dots)$
 
is the second one in Y?
 
(a,a,b,b/2,c,c/4,etc), no?
 
ah!
 
9:06 PM
Or is this not zero-indexed
 
I misunderstood :P
never mind me
 
Okay so let's say we have $\mathbb{RP}^n$ as $S^n$ mod antipodal points, I want to see why this is $D^n$ with antipodal points of the boundary identified.
 
Hemisphere.
 
is it?
D^n has trivial homology
 
It is
 
9:11 PM
Hatcher says so, which throws me off a bit because $S^n$ already identifies all the points on the boundary of $D^n$
 
$D^n$ is homeomorphic to the hemisphere.
The equator is its boundary.
 
If S^n is thought of as the one point compactification of R^n, even better D^n is defined as the hemisphere
I guess I'm implicitly saying that the lower hemisphere is taken to the unit ball under stereographic projection
 
Okay that makes sense actually, so then you take a point from each equivalence class in $S^n$ except you take both points on the equator/boundary of $D^n$, so you identify them
 
A point in $S^n/{(p\sim-p}$ is a pair of opposite points. Either both points are on the equator, or one is in the open southern hemisphere and the other is in the open northern hemisphere
 
wait, isn't S^n just D^n quotient boundary?
 
9:14 PM
You can forget about the open northern hemisphere 'cause it's extra information
 
Okay so I'm now happy with the CW structure of $\mathbb{RP}^n$
 
and you probably have to talk about how $D^n$ (the hemisphere) is a closed subset of $S^n$, now that I think about it
so that it ends up with the right topology
'cause otherwise you could say the same argument about $D^n$ minus half its boundary
 
9:31 PM
$(-1)^{1/\log(x)}$ can be written as $e^{i\pi/\log(x)}. can for example, $(-1.32)^{1/\log(x)}$ be written in a similar way
 
You could also construct a projective plain by taking three squares and four triangles and arranging them in this self-intersecting way
Imagine an octahedron
 
@Ultradark $\log(-1.32)=\log(-1)+\log(1.32)=i\pi +\log(1.32)$
 
Delete half the faces (i.e. imagine coloring the faces in two colors in a checkerboard fashion and delete the faces of one color)
 
so therefore $-1.32=e^{i\pi+\log(1.32)}$
 
You know how an octahedron is two square pyramids stuck together? So it's got a square in it
 
9:34 PM
and then (-1.32)^(1/log(x)) = e^(i*pi/log(x)+log(1.32)/log(x))
 
Three, actually, if you rotate it
Put the squares there
This gives you a polyhedral, self-intersecting model of the projective plane
(The double-cover is isomorphic to a cuboctahedron)
I imagine you could probably smooth this out somehow
Yeah
 

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