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

12:01 AM
@leslietownes. Thanks Prof

maybe less formally, if a and a' are different elements of the domain of f and f(a) = f(a') = b, then there's no one single choice of what f^{-1}(b) ought to be. there are at least two possibilities. if f^{-1} were a function there would be only one.

Many English words end in 'ck', but none end in 'kc'
With your help, we can change this

how about a replacement schema where words ending in ack, ick, ock, and uck are all changed to end in -kc instead of -ck? then we could have a mnemonic of the form 'k before c except after e'

12:34 AM
Good evening everyone. I'm still having fun messing around with function approximation.
It seems there is a general method of approximating a function by taking an approximation of the function and either the exact inverse or an approximation of the inverse. Is there anything in the literature about this?
I assume there is

at that high level of generality it seems more of a vibe than something people would write papers about. but yeah, i could imagine stuff like that. if for some reason some number a is thought to make f(a) resemble f(x), and some g is thought to resemble f^(-1), maybe try g(f(a)) as a replacement for a in an iterative scheme to find things that will give you f(x).
anything resembling literature that i've seen on this is far more specific in its hypotheses, but i could see the vibe.

12:52 AM
Well unfortunately I can't claim some "function of function approximations" as my results currently are subjective and require adaptation based on circumstances, but what remains objective for the time being is that it works for finding an approximation of a function and its inverse solely by computing the error to x itself.
I'm currently trying it with $2^x$
I did however try just the other day to see what would happen if I put different functions other than x into the recursive reciprocal function that I found. Curiously, for certain domains of x, it converges to the reciprocal of the function I put into it, whereas others went off to infinity.

1:54 AM
how to write $f(x,y,z) = xy^{2} + \sin(xz) + e^{z} - 2 = 0$ as a function of $(x,y)$ near $(1,-1,0)$? I established it can be done.....but doing it...well nothing "nice" is happening.....

After some playing around, I got things "to work", but I only know it works because I have a solution....I don't understand why it failed for one scenario but not for another.
After having established $f(x,y,z) = xy^{2} + \sin(xz) + e^{z} - 2 = 0$ can be written in terms of $x,y$, looking at the $\sin$ term I "bounded" my expression: $xy^{2} + e^{z} - 3 \leq xy^{2} + \sin(xz) + e^{z} - 2 \leq xy^{2} + e^{z} - 1$
I was asked to find solutions for the partials of $z = \phi(x,y)$ w.r.t $x$ and $y$ at the point $(1,-1)$. Initially I used the upper bound and got the incorrect answer, but using the lower bound I got the correct answers. Both produced simple values. Is there some reason why only the lower bound solutions are shown?

2:13 AM
avra, happy. she just ate a pretzel.
dc3rd, were you asked to do this? a lot of implicit function theorem stuff just leverages existence of the function (which you can establish by checking derivative conditions) and the functional relationship. the example may have been chosen precisely because it's difficult to 'solve for z' in any useful sense.

Just for clarity after using one of the bounds I rewrote things as $xy^{2} + e^{z} - 1 = 0 \Rightarrow z = \ln(1 - xy^{2})$
I was asked specifically to find $\frac{\partial \phi}{\partial x}$ and same for $y$ at the given point $(1,-1)$. I wasn't asked to solve for the $\phi$ explicitly though. but I couldn't see any other way to compute the derivatives. And yes you're right there is no "easy" way to solve for $z$ in my expression....

is the point to identify the extent of some domain around (1, -1) where z is a ufnction of x, y? that must be what this bounding stuff is about??
not sure?

The bounding stuff is me being maverick about things and trying to find some way of writing $z$ as a function of $(x,y)$.

there's a thing about being a maverick. you're on your own. that's why you're a maverick!
i'll be over here eating the 1/4 of a pretzel my daughter left me

how the question was asked was to establish that I can write $z$ as a $C^{1}$ function $z = \phi(x,y)$. then the second part asked me to just compute the partials at the given point. So I thought the only way that can be done is by finding said function of $z$, but as you saw it is not a fun endeavor....

2:23 AM
oh yeah, there's an 'implicit' way to do that. may help to think about how it works in one variable. it's the same.

hmmm....ah yes I looked over the clearly marked lemma that I chose to just glimpse at. in particular

$$\frac{\partial \phi}{\partial x_{j}}(\mathbf{a}) = \frac{\frac{\partial f}{\partial x_{j}}(\mathbf{a})}{\frac{\partial f}{\partial x_{n}}(\mathbf{a})}$$
forgot the minus sign

2:56 AM
Folks, Writing $1/2\sqrt{2}$ or $1/2x$ is ambiguous or not?

ambiguous

@dc3rd OK. Thanks.

2 hours later…
4:57 AM
always nice when a simple question has a crisp, right answer.

For every positive sequence $(a_n)$, it is to be proven that $\limsup \frac{a_1+a_2+...+a_n+a_{n+1}}{a_n}\ge 4$

the lim sup bug has really bitten you hard, hasn't it

:)
Let $x=\limsup \frac{a_1+a_2+...+a_{n+1}}{a_n}$ . Let $\epsilon\gt 0$ be given. For large $n$ we have $\frac{a_1+a_2+...+a_{n+1}}{a_n}\lt x+\epsilon\implies \frac{a_n}{a_1+a_2+...+a_{n+1}}\gt \frac 1{ x+\epsilon}\implies \frac {a_n}{a_1}\gt \frac 1{x+\epsilon}$

my cat is so close to 500 followers on instagram. i know most of them are bots but i don't care. it feels like it matters.

more followers than I ever had when I had an IG account...I was probably missing "provacative" pics and captions tho....

5:12 AM
black cat instagram is a pretty close-knit community. there's a tiny fraction of people who only want to post and see pictures of black cats, and we all follow each other.
then there's a broader population of casuals who come and go.

Using the other implication of limsup definition, i.e. there are infinitely many $n$ such that $\frac{a_1+a_2+...+a_{n+1}}{a_n}\gt x-\epsilon$ but here the problem is RHS could be zero so I think it doesn't help much :'(

one time i was interviewing someone for an internship and we had twice as much time because the previous interviewer had to run at the last minute. we got to talking about instagram, including black cat instagram. it turned out that they followed my cat already. i didn't let this affect my decision on the interview.
they got an offer and turned it down, which is a shame.

the possibilities for collaborative black cat posts all gone up in smoke

i wish to adopt a cat one day.
and a dog also

koro, that's a good goal. for a long time i lived in places that wouldn't have allowed a cat. or i couldn't have afforded it.
having a cat now is a very nice thing. despite my complaints.

5:21 AM
one of my relatives has two Persian (one white and the other like Garfield) cats. :)

they must have fun brushing that hair off of everything. :)
olivia has short hair and i mostly wear dark colors, so it's mostly fine if my clothing is covered in cat hair. even my suits.
i'm not sure that white persian fur would be as compatible with my wardrobe.

But does Olivia shred hair?

certainly not like a persian does. but if she sits in my lap or i pick her up and hold her for a while, she will leave a tiny bit of fur.

I mean is it too much to clean up hairfall? I never had a pet, that's why I ask.

with shorthair cats it's pretty easy. it helps if you brush them. that reduces shedding. you can avoid brushing them, but you may need to sweep up every week or two.
and keep a brush or lint roller around to get hair off of strategic items of clothing.
i swept around my desk today for the first time in a few months, and most of what came up was black cat hair. it wasn't a lot, but it was there.
i'd probably notice it more if it were a bright white color. as it is, i never notice it.

5:28 AM
hi, apparently the following is an easy result, but Im not seeing how to prove it: Let $u$ be harmonic in $\mathbb{C} - \overline{\mathbb{D}}$, such that we know $\partial_{r}u = 0$ on the boundary of the unit disk (and we know $u$ actually has a harmonic extension to some small neighbourhood of the boundary where this partial derivative is defined, say), then $u$ extends to the interior of the unit disk , by defining it as $u(\frac{1}{\overline{z}})$ in the interior
any takers?

is this something that just drops out of formalism with d-z and d-zbar

could be
hmm
whoops! I meant interior minus origin
well okay, for sure that $u(\frac{1}{\overline{z}})$ is harmonic in the interior minus origin is just a computation and yes you can just use dz,d-zbar to make your life much easier and see it very quickly

i learned most of what i know of complex analysis from a PDE guy who would just do that. he was a magician.

ugh, i forgot the most important part of the question
by 'extends to the interior of the unit disk', I mean, since we already know $u$ is harmonic in a neighbourhood of the boundary, that it extends to the punctured plane.
that is, yes $u(\frac{1}{\overline{z}})$ is easily seen to be harmonic in the interior minus teh origin
but we need to show that this also gets you harmonicity at the boundary
hopefully I am making more sense now
and this is certainly where we will need to use more information about $u$ on the boundary, in particular that $\partial_{r} u = 0$ there

where we're mixing and matching the formulas. that makes sense.
is this an exercise in ahlfors? i think i've seen it before. or not. i may be blending this with hazy memories of stuff like the schwarz reflection principle for analytic functions.

5:35 AM
its not an exercise in ahlfors
but its definitely related to the reflection principle
reflecting with respect to $S^1$ is exactly $z \rightarrow \frac{1}{\overline{z}}$

maybe it's in conway. except i don't remember conway spending a lot of time on harmonic functions.
hrm.
this is where i dig in my notes to see if i can just cut-paste an answer.
in the schwarz context i think one way of proving the analyticity in regions containing the reflection boundary is using something like morera's theorem. you sort of sidestep the whole pointwise thing, and decompose into averaged behavior. is there something similar we can do for harmonic functions?
averaged behavior might not be the right word. you replace emphasis on point values with emphasis on values along curves.
something like the mean value property maybe?

yeah there is a similar thing, in fact rather than using moreras theorem you can prove a stronger version of the schwarz reflection principle by starting with some harmonic function $v$ that is define in the upper half plane and continuous at the boundary (where it vanishes), then define it in the lower half plane by $\overline{v}(\overline{z})$ and the mean value property now obviously holds at the boundary because the integrals over the upper and lower semicircle are equal in magnitude
(with opposite signs)

ok, so we are at least on the same page even if we're no closer to the result.
hrm.

the issue with this is im not sure how to relate $\partial_{r}$ to the behaviour over some small circle

does that let you sneak up on points on the boundary with curves that get close to but do not touch the boundary?
hrrm.
my advisor used to 'hrrm' and i acquired this as a nervous habit. trust me, it works.

5:49 AM
haha, i do the same
it puts me at ease even if ive made no progress, lol

hrrm

i still hear it in his voice in my head.

by the way, just to relate this even more with the schwarz reflection principle, let $z$ be a point on the boundary and take some small disk around it where $u$ is still harmonic (i mentioned earlier its harmonic in a small neighbourhood of the boundary), it has a conjugate $v$ in this disk, so that $u + iv$ is analytic, now $\partial_{r}u = \frac{1}{r} \partial_{\theta} v = 0$, so $arg v$ is constant on the boundary arcs
uh, well that doesnt relate it the schwarz actually, sorry, but it may still be useful
wait.. no, $v$ is constant on the radial lines..
forget what I said about $argv$

what else are you lying to me about
:)

nothing I promise :D

6:00 AM
there's gotta be something dumb going on here that i'm just missing

indeed
im sure this is true because it was told to me by someone who is a pretty renowned potential theorist/complex analyst
that being said, this person said after I asked if he had a reference for the result 'no, but it is easy to prove'
maybe it is easy for him to prove

i wish i had the authority to say stuff like that. in my job, this is reserved for judges, who can say whatever they want and it becomes true.

Let $x:=\limsup \frac{a_1+a_2+...+a_{n+1}}{a_n}$ . Let $\epsilon\gt 0$ be given.

If $x$ is infinite then there is nothing to prove so let's consider the case when $x$ is finite.

For large $n\ge K$ we have $\frac{a_1+a_2+...+a_{n+1}}{a_n}\lt x+\epsilon\implies \frac{a_n}{a_1+a_2+...+a_{n+1}}\gt \frac 1{ x+\epsilon}\implies \frac {a_n}{a_{n+1}}\gt \frac 1{x+\epsilon}\implies \frac{a_{n+1}}{a_n}\lt x+\epsilon \implies \limsup \frac{a_{n+1}}{a_n}=x.$
But how can this help me conclude that $x\ge 4$ for every positive sequence $(a_n)$?

3 hours later…
9:22 AM
Say I have a covariance matrix $C$, since it is symmetrical, its eigendecomposition is $C = W \Lambda W^\intercal$.
Since it is the same form as the SVD $C = U \Sigma V^\intercel$, can I argue that for a symmetrical matrix, its SVD is its eigendecomposition?

1 hour later…
10:46 AM
@PeterJohn I saw your post. And in Google, there is a paper about that problem which proves not for general topological space but for metric space. I'm not sure the problem statement is true for general space. Why don't you try first for simplicial complex?
And I'm also quite sure join operation is irrelevant

11:38 AM
Can anyone give me a hint on how to integrate this: x/x+1 ? I don't understand why I cant treat it as x â€¢1/x+1 and go from there.

Hello, can someone explain to me how this user did his contour integration ?
https://math.stackexchange.com/a/2372334/272494
If we want to integrate from on $[0,-iR]$, then this is the same as integrating on $[0,R] \cup \{Re^{i \theta} : \theta \in [0,\pi/2] \}$. But then I don't understand how he obtains his result.
I think I got it. I gave up too fast :)

12:22 PM
@Jimbo if youâ€™re not comfortable with the algebra, you can instead start from the substitution y=x+1
then x/(x+1)=(y-1)/y=1-1/y

1 hour later…
1:38 PM
2

A circle is exactly defined by three distinct non-collinear points. But I need a way to solve the following problem (all in 2D): Given three points, calculate a circle with all three points on its border if it exists, else calculate a circle with minimum radius which has two points on its border...

1:53 PM
If I had a joint pdf $$f_{X,Y}(x,y)$$ and I then wanted to transform it using the function $U = X/Y$ would I still use the Jacobian method - but I don't have a second transformation. Would I just define some random R.V and continue as normal?
any ideas?

2:13 PM
@huzaifaabedeen You have posted that question three times in chat this week. Semiclassical posted an approach. You have not posted any thoughts or responses. Why?

Hi professor Rob!

@huzaifaabedeen You've even posted a bounty. Why are you continually posting it to chat?

I have one small question that's been bothering me since today's noon.
I have $\limsup x_n=x$ then I want to show that there exists an $N\in \mathbb N$ such that for all $n\ge N$, it follows that $x_n\le x$.
This seems correct when I tried examples of sequences $(x_n)$ and $(y_n)$ where $x_n=1$ when $n$ even and $x_n=2$ when $n$ odd; and $y_n=0$ when $n$ odd and $y_n=-n$ when $n$ is even.

@Koro That's false
What's limsup x_n where x_n=1/n?

I know that by definition of limsup, we can say that given any $\epsilon\gt 0$ there exists $N$ such that $n\ge N\implies x_n\lt x+\epsilon$
@AlessandroCodenotti $0$ ... Oh my God.
I had a feeling that the result was false but I couldn't come up with a counterexample.
This means that the last part of this answer (math.stackexchange.com/a/1863959/266435) is wrong.
right?

2:24 PM
@Koro Yes, this has been pointed out and fixed in the comments

@AlessandroCodenotti thank you so much for your help. I have also commented there based on nice counter-example suggested by you and tagged Mr. MartinR with whom I discussed this question (math.stackexchange.com/questions/4289711/…) today.

Hi @Alessandro

Hi @Balarka how are you doing?

I had a couple of observations I wanted to pass you by

On which topic?

2:33 PM
Dimension theory

You have my attention

So there's this result by Hurewicz which says that if $f : X \to Y$ is a map (say, between compact metric spaces, although I am sure this can be relaxed) then $\dim X \leq \dim f^{-1}(y) + \dim Y$, yes? The equality fails by Pontryagin's example.

Right (and yes it can be relaxed a lot but I don't remember the exact assumptions, it's some technical general topology, for sure it is written in Dimension Theory of General Spaces by Pears)

Great. So I believe that in my setup, you have $\dim X = \inf_{\varepsilon} \sup_{\mathcal{U}} \dim N(\mathcal{U})$ where $\mathcal{U}$ runs over all covers of Lebesgue number $\varepsilon$ in the inner sup, and $N(\mathcal{U})$ is the nerve, and by $\dim N(\mathcal{U})$ I just mean the dimension of the simplicial complex. Obviously, this is bounded by the least $n$ such that $\check{H}_c^{* > n}(U)$ vanishes for all open subsets $U$ of $X$. If I understand correctly, these are actually equal?

@robjohn I am sorry robjohn Ididnt see semiclassicals reply yesterday

2:40 PM
$\check{H}_c^*$ denoting the compactly supported Cech cohomology in degree $*$

Hm, not sure, I never looked in detail at dim in terms of Cech cohomology

I think this is a theorem by Alexandrov (Alexander??)
I understand one inequality, because $\dim N(\mathcal{U}) = \dim \check{H}(X, \mathcal{U})$

(so it seems that the most general result is with T_4 X, weakly paracompact Y and a continuous closed surjection f:X->Y, and you get $\dim X\leq \mathrm{Ind}(Y)+\dim f$ where dim f is sup of dim of fibers)
When Y is metrizable Ind is strenghtened to dim

The reason I ask is I think the analogous Hurewicz theorem in Cech cohomology can be proved with general nonsense. Given a map $f : X \to Y$ of locally compact Hausdorff spaces, there is an isomorphism $\check{H}^\star_c(X; \Bbb Z) \cong \check{H}^\star_c(Y; Rf_!\Bbb Z)$ where $Rf_!\Bbb Z$ is a sheaf of complexes, whose stalks have homology $\check{H}^\star(f^{-1}(y); \Bbb Z)$.

Yes one inequality is the fact that Cech cohomology vanishes in degree above the covering dimension (which is false for say singular cohomology)

2:49 PM
Right.

@BalarkaSen Oof I know way too little about both sheaves and Cech cohomology to comment

Think of it like a twisted Kunneth theorem. $\check{H}_c(X) \cong \check{H}_c(Y) \widehat{\otimes} \check{H}_c(f^{-1}(y))$ where $\widehat{\otimes}$ indicates some kind of twisted tensor product
This is general nonsense, you don't have to know the proof. It's not geometry
But it should tell you that the Cech cohomological version of dimension satisfies the Hurewicz inequality
More precisely it should give $\dim X \leq \inf_{y \in Y} \dim f^{-1}(y) + \dim Y$

inf seems extremely suspicious, with a sup it seems reasonable

Is that so? Don't you have $\dim X \leq \dim f^{-1}(y) + \dim Y$ for all $y \in Y$ for covering dimension?
And then you take inf on both sides

No, you take sup over y in Y of dim f^-1(y)

2:54 PM
Ah OK
You may be right

I can always take disjoint union of X and Y with a point and send the new point to the new point to artificially introduce a zero dimensional fiber without affecting the dimension of X and Y

Yeah, of course that's correct.
I spoke without thinking. In my $\check{H}_c(X) \cong \check{H}_c(Y) \widehat{\otimes} \check{H}_c(f^{-1}(y))$, the second component is in fact varying point to point wrt $y$
I can only bound by the largest dimension, that's right
I haven't checked all the details here but I feel like this should give a proof of Hurewicz theorem. My ultimate question is if a similar proof exists for the Bell-Dranishnikov's Hurewicz theorem for asdim
There's an analogue of this Cech stuff, where you use bounded cohomology, by Dranishnikov.

Ah yes I have vague memories of looking at that while writing my masters thesis

I really feel like that'd be a good proof
I asked Bestvina during a minicourse he gave here and he said that seems reasonable.
I don't know enough about asdim so I am hoping you'll write it up instead lol

lol I don't know about the cohomological approach either
Also I realized I don't remember much about asdim either. I was asked the other day whether there are groups that have exponential growth, finite asdim and are amenable
I'm not sure whether that is possible though

3:06 PM
What's the asymptotic dimension of the lamplighter

That's what I thought about too but I'm not sure

It feels like 1 lol
Forget all the lamps

@love_sodam Any difference when we assume $X$ a simplicial complex?

balarkaaa

Hey @fin!

3:12 PM
.do you remember me

Indeed I do

yayyy

@BalarkaSen Yeah that's a reasonable guess

i need helpp

@BalarkaSen that's pretty much the proof that the group is one ended lol

3:13 PM
Ahh @AlessandroCodenotti
@fin Go on

ok so i need to take a line integral of a vector field along a certain path
my book says thats done by integral of v dot dl
where dl is an infinitesimal displacement vector
right

yeah

how do i calculate dl? im sorry this is a stupid ass question but im not good at math
and its been a while
because i know v is just the function i was given

@PeterJohn Since $X$ is a free $\Bbb Z_2$-space, there is a natural partition of $k$-dimensional simplices in $X$ for each $k$.

also if we take a line integral of the same path but its parametrized to be slower along time would i still get the same value?

3:17 PM
If you wish, $d\ell$ is the infinitisimal tangent vector to the path

how do i know the dimensions though

@fin Yeah. That's a little surprising, right?

@BalarkaSen a little bit but it makes sense because the faster you go the bigger d\ell is
or not

Yeah, which is why $d\ell$ is the tangent vector to the path

like for example one of the paths i have is from (0, 0, 0) to (1, 1, 0)
is $d\ell$ equal to (dx, dy, 0) then?

3:19 PM
One moment. Your path is the straightline $(t, t, 0)$, $0 \leq t \leq 1$?

yes

Then it should be $(dt, dt, 0)$, yes?

ohhh wait
the thing thats confusing me is that the book im using is doing it in shorthand
like he doesnt use any mention of $t$

Tell how they're doing it
It's one of the awful things about engineering calculus.

lemme get a pic
yeah its really confusing
i thought it was just me
but like theres so much shorthand and its really confusing

3:21 PM
Nope, I have been there

i recently learned engineers use integration tables
well, it's not so bad since actually I think you'd expect most relevant integrals relevant to engineers to be solved by computer nowadays

3:42 PM
@love_sodam That's nice but still, I have no idea

@fin Ah OK I see.

so wait
is it ok if i do it the parametrization way
or does this book want me to do it their way
because the shorthand notation really fucks w my intuition

OK, so here's the difference between what you're doing and what I'm doing. When you write $d\ell = (dx, dy, 0)$, you're writing the tangent vector to the path at the point $(x, y, 0)$ in the coordinates of the 3d-space:
$de_1 = (dx, 0, 0)$ is the infinitisimal direction in the $x$-axis, $de_2 = (0, dy, 0)$ is the infinitisimal direction in the $y$-axis, $de_3 = (0, 0, dz)$ is the infinitisimal direction in the $y$-axis. So the infinitisimal direction in the $z = 0$ plane in the $x = y$ direction is $de_1 + de_2 = (dx, dy, 0)$.
I much prefer using parametrized, where I use the fact that the path $\gamma(t) = (t, t, 0)$ has intrinsic 1D coordinates since the domain of the path $\gamma$ is a 1D space $\Bbb R$.
In that case, I just view $d\ell$ in the intrinsic coordinates, so it becomes $"d\ell =" \gamma'(t) dt$ where $\gamma'(t)$ is the genuine derivative of the path at the time $t$
And $\int_\gamma F \cdot d\ell = \int_a^b F(\gamma(t)) \cdot \gamma'(t) dt$
This is called "pullback" in differential geometry lingo. I pullback $d\ell$, which lives in $\Bbb R^3$, the target of $\gamma$, to the domain $[a,b] \subset \Bbb R$ of $\gamma$
If you don't care about the formalism, you can take $\int_\gamma F \cdot d\ell := \int_a^b F(\gamma(t)) \cdot \gamma'(t) dt$ as the definition of line integral of a vector field $F$ along the path $\gamma : [a, b] \to \Bbb R^3$.
It makes sense also, right?
I recommend doing it this way because 1 coordinate is superior to 3 coordinates, and you don't have to bother with what algebraic manipulation with infinitisimal quantities like $dx, dy, dz, dt$ etc all these mean rigorously

i also recommend doing it that way, if only because i've always done it that way

3:56 PM
@BalarkaSen do you think its okay if i do line integrals this way?
bc its different from the book but if im getting the right answer it shouldnt matter right
?

what course is this for

electrodynamics

No way they'll object
Universally accepted as the right formula by physicists

side note, balarka, stop revealing the hidden secrets of the mathematicians

@BalarkaSen also if it was going to $(1, 2, 0)$ instead would be $d\ell = (dx, 2dy, 0)$
and like whats stopping me from doing $d\ell = (dx/2, dy, 0)$?

4:04 PM
If $X\times I$ is embedded in $\Bbb R^n$, must $X\times\{0\},X\times\{1\}$ be ambiently isotopic to each other? Or does $X$ need to be compact
oh wait I think I see a noncompact counterexample
I assume it's true for compact $X$

@fin You're scaling the actual unit tangent vector, which is $\gamma'(t) = (t', (2t)', 0') = (1, 2, 0)$ down. But try to compute line integral of $F(x, y, z) = x^2$ along two different parametrizations of the straightline joining $(0, 0, 0)$ and $(1, 2, 0)$ given by $\gamma_1(t) = (t, 2t, 0)$, $0 \leq t \leq 1$ and $\gamma_2(t) = (t/2, t, 0)$, $0 \leq t \leq 2$.
Use the good definition

i mean i believe itll give the same answer but i just feel bad because i dont understand the way the book does it
which makes me think i dont know my calc as well as i should lol

the book is doing it in a stupid way, it has nothing to do with how well you know calc
once you compute enough examples using the good definition you'll understand the bad definition as well

I have a question about naming variables. If N and N are two unrelated variables but I want to use the same N for both of them. Should I use: N' or \hat{N}

Does $X$ need to be connected...?

4:09 PM
@AkivaWeinberger I don't think so. There's an embedding of $S^2 \times I$ in $\Bbb R^3$ such that at time 0 it's the standard sphere and at time 1 it's the Alexander horned sphere, I think

hi akiva remember me

what is the convention for naming variable \hat{N} vs. N'

@fin not sure...?

meow mix

Ah! Yes

4:10 PM
im a freshman in college now lol

oh wow i thought you're 13 or something

Where?

ramapo college lol

nice

@BalarkaSen i was when we last spoke probably lol

4:12 PM
No, when we last spoke you were drunk.

idont remember that

I remembered I was just ribbing
Of course not

in fact im at the liquor store rn

Jeez.

i work there silly

4:14 PM
Aha

@AkivaWeinberger Not in a stencil ;-)

ok wow this good definition makes everything a LOT easier

Told ya

albeit a bit more tedious but i didnt expect anything else from vector calculus

hello, please if f is strong continious then f is weak continuous ?

4:19 PM
@BalarkaSen Wait crap you're right

Indeed.

It's just the solid horned sphere minus a small ball

Isotopy extension fails badly in topological category

'cause the solid horned sphere is a ball

That's wight, wabbit

4:20 PM
What about $S^1\times I$

Sorry
Probably wild arc nonsense
It should even fail for $I \times I$
Try the Fox-Artin

Not sure I see it
Wait
Yes I do
Oh my god

All your life is a lie because you didn't know F-A are actually isotopic to trivial but not ambient isotopic to trivial?

Take the Fox-Artin horned sphere ("thicken" the Fox-Artin wild arc so that the amount of thickening goes to zero fast enough near the endpoints). Its interior plus boundary should be homeomorphic to a ball
and then you can embed IxI into any ball where one edge is anything completely inside and the other edge is anything on the boundary

Yeah that's what I had in mind

4:25 PM
In fact
Thicken F-A into a ribbon
Voila, embedding of a rectangle
(aka a disc)
We can take any two segments of the boundary of a disc to be opposite edges of IxI

Smart

OK what about $S^1\times I$
Hopefully I can't have a trefoil on one end and an unknot on the other

It's true for smooth embeddings
That's the smooth isotopy extension theorem
I mean why are we working this hard just take any knot and then shrink a part of it until it goes to 0

Sure but even if the edges are smooth the interior might not be
@BalarkaSen Not sure I see how to do that so that the whole cylinder stays embedded
It can't intersect other timestamped versions of itself

Sorry, but please if f is strong continuous can we deduce that f is weakly lower semicontinuous ?

4:30 PM
You can do it in $\Bbb R^{100}$

@Vrouvrou I continue to not know functional analysis, sorry

Oh one second I forgot the question
In R^100 there's no such thing as knots.

You can of course make part of a circle a Fox-Artin wild arc
But you want nice things
Sorry, what do you want, actually?
You want to know if there's an embedding of the annulus in R^3 in a way that the base is trivial and the top is a trefoil?

vrouvrou, what are the domain and codomain of f and what are the notions of strong and weak lower semicontinuity? at a high level, intuition suggests 'yes' because strong beats weak
but that's the law of the jungle and not math

4:40 PM
@Vrouvrou if we assume what you're asking is true, then if we have a functional $f:X \to \Bbb R$ that is strongly continuous, then $-f$ is strongly continuous as well. So both $f$ and $-f$ are weakly lower semicontinuous, but that means that $f$ is both weakly lower semicontinuous and weakly upper semicontinuous, which implies that $f$ is weakly continuous. But not every strongly continuous functional is weakly continuous
@leslietownes terminology is a bit confusing here, but actually for a functional on (say) a locally convex vector space weak continuity is stronger than strong continuity

@AkivaWeinberger I mean if you could then you could cap off the base and get a disk bounding a trefoil; the trefoil has knot genus 1.

$f: W^{1,p(x)}_0\to\mathbb{R}$

yeah, i never know what to expect. i can think of settings where this might be true. but i don't want to assume.

Genus 0 knot iff trivial

f is strong continuous means $u_n\rightharpoonup u\Rightarrow f(u_n)\to f(u)$

4:44 PM
@BalarkaSen Not necessarily, the base might have to intersect the cylinder

and weak continuous $u_n\rightharpoonup u\Rightarrow f(u_n)\rightharpoonup f(u)$

@Vrouvrou isn't that the definition of weakly continuous?
@Vrouvrou these two definitions are the same! a sequence in $\Bbb R$ converges weakly iff it converges strongly

@AkivaWeinberger You can trap the annulus away from a collar nbhd of the base in a big bounded set, and then extend the base to infinity and cap it off at $\infty$ in $S^3 = \Bbb R^3 \cup \infty$

so for a functional strong=weak ?

@Vrouvrou no

4:46 PM
Oh I guess maybe the base and the top could be linked

Actually, the definitions as I know them:
$f$ is strongly continuous if $u_n \to u \Rightarrow f(u_n)\to f(u)$ (that's just the usual continuity)
$f$ is weakly continuous if $u_n \rightharpoonup u \Rightarrow f(u_n) \to f(u)$
@Vrouvrou that just means that for a functional $f$, it doesn't make a difference if you write $f(u_n) \to f(u)$ or $f(u_n) \rightharpoonup f(u)$. But $u_n \to u$ or $u_n \rightharpoonup u$ still makes a difference

@Akiva: OK, cut the annulus along a height. This gives you embedding of a disk with boundary $K \# Trivial = K$

i understand @LukasHeger

@BalarkaSen No they still could be linked
Like, uh, you know how you can embed a cylinder where both edges are a trefoil? By ribboning it?
Cut that along a height. We know K#K doesn't equal O
I think our best bet is showing the cylinder can be "smoothed" somehow and then use the smooth extension thingy
'cause the boundaries are already smooth
and everything else "looks" like the interior of R^2 intrinsically

Yeah but the interior is messed up so I thought approximating by smooth maps won't make it an embedding anymore

4:55 PM
Hm I guess we don't actually know the map is smooth near the boundaries, even if the boundaries themselves are smooth

@AkivaWeinberger The point is the tube you're using to take $\#$ is messed up?

For # to work there has to be a sphere intersecting the result at only two places

Oh yeah that's right

so a sphere separating the summands that the tube hits nicely

@LukasHeger if it is weakly continuous then it is weakly lower semicontinuous

4:59 PM
@Vrouvrou yeah

thank you very much

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