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16:00
So given a loop $[0,1]\to S^2$ can you always find a point in $S^2$ outside of its image for this argument to go through?
Can you prove it?
hahaha no
i have to talk about separating the sphere
in the interior and exterior
with a curve
Well think about it then
or something
16:02
there are space-filling curves, so this is not obvious
oh you are saying that there are loops that will cover the whole sphere so i cant find a point to do the above proof
also if i consider the loop
through the north pole
sperating the sphere exactly at 2 hemospheres
There are no loops that do that, but it's not obvious
how can i say if the point is the interior of the curve or not?
is in*
not sure what you mean by interior on a sphere
@MatheinBoulomenos Are you sure about that?
you should really talk about the image of the loop rather than the interior, a loop is a function $[0,1]\to S^2$, not a subset of $S^2$
16:07
@MatheinBoulomenos interior of a curve i mean the set that has boundary the curve
ohh i thought we said that if the point is not in the image of curve but it is the interior we have a problem
ok so no interior is involved after all just the image
hm.. still i cant think why i cannot always find a point not in the image of the loop
Hmm, I can only prove this if $S^1 \to S^2$ is injective
which doesn't say much
I'm pretty sure that there are such loops
we havent talked about space filling curves or anything like that
FInd the Error of this Proof : $S^2$ is simply connected: Proof. Let a loop on $x_0$ of the sphere and take a point $p$ in the $S^2$ also not in the loop. Now $S^2 \setminus p$ is homeomorphic with the plane and tha plane is simply connected so the loop is homotopic to a point hence $S^2$ is simply connected
"take a point $p$ not in the loop"
Anyway, the only step where the proof can go wrong is if you can't find such a point $p$. You don't actually need to know whether there are loops covering the sphere to fix the proof though, suppose you have such a loop and you can't pick $p$, modify the argument so that it works anyway
16:13
whether that's true or not, you have to justify it
Just cover $S^2$ by $S^2\setminus \{p\}$ and $S^2 \setminus \{q\}$ for two different points and apply Seifert-Van-Kampen
@AlessandroCodenotti how exactly would you do that?
yeah in dont need to prove that the sphere is simply connected
i need to find an argument on how the proof is wrong
@MatheinBoulomenos I did that by showing that every loop is homotopic to a nonsurjective one, but there might be different approaches
so the proof is right?
just need to prove that every loop is not surjective?
and so there exists such a point ?
There are surjective continuous functions $[0,1] \to [0,1]\times [0,1]$, so there is no reason to expect that loops can't be surjective
There are such loops
I think
at least there are surjective continuous functions $[0,1] \to S^2$
@MatheinBoulomenos Call such a function $f$, since $S^2$ is a quotient of $[0,1]^2$ isn't $\pi\circ f$ such a loop?
Ah, wait, we don't have $f(0)=f(1)$, but that doesn't seem problematic
16:19
@AlessandroCodenotti the space-filling curves I know are not loops
That one should do the trick
yeah i was looking at that one
okay, there's the answer
so the answer is: The proof is wrong if the loop is space filling?
16:22
Yes, because you can't pick a point $p$ as described
hm.. weird i dont think our teacher was expecting us to know the existance of such curves especially since he hasnt mentioned anything in the notes
might there be a simpler argument
However if you can show that every loop in $S^2$ is homotopic to a nonsurjective loop you can use that proof to show that $S^2$ is simply connected (Of course just using SvK is much easier)
There are surjective loops on every connected, locally connected second-countable compact Hausdorff space
this follows from the Hahn–Mazurkiewicz theorem and the fact that there is a continuous surjection $S^1 \to [0,1]$ (just take a projection on a coordinate and rescale)
16:25
That's a cool theorem I've never heard before
are you guys students?
yes
undergrad student in my 3rd year
16:26
really?
damn you guyz are good
@ManolisLyviakis yeah
im undergrad too xDD on my last year
The 3rd is the last undergrad year here, I'm not sure how it works elsewhere
4 years in greece
and it can be extended to 6
The recommended time to finish is 3 years, but a lot of people take longer (3,5 or 4 years is pretty normal)
16:28
if you not that good and dont pass your exams
our average is doing 5 years here
@MatheinBoulomenos followed by 2 more years for a master?
at least 2 years, yes
it's the same for the masters
some people take 3 years
Yeah, that's how it works in Italy too
I won't finish in 3 years and I didn't fail a class. That's my fault for taking only classes I'm interested in and delaying the boring stuff
same here
im on my 6 year im 24 years old xD ive delayed my aplied math courses really hard
i blame you on my anxiety now hahaha
16:33
what did I do?
I should finish on time in July
@Alessandro do you know where you will do your masters?
I'm thinking in Turin, Italy
They have a cool logic and set theory program
And there are a lot of free credits so I can also do plenty of geometry courses
sounds good!
I'll probably won't do much algebra and I'm trying to stay as far as possible from analysis :P
16:37
not doing algebra doesn't sound so good, but not everyone needs to do the same
38 mins ago, by Alessandro Codenotti
So given a loop $[0,1]\to S^2$ can you always find a point in $S^2$ outside of its image for this argument to go through?
i took my first course on topology this semester and fell in love but i find it really hard
why would the domain be $[0,1]$ @AlessandroCodenotti
@ManolisLyviakis hard as in requiring formalism?
because that's how a loop is defined
I grew to hate the $[0,1]$ idea. Why don't you just define the domain to be $S^1$
If you say a loop has domain $[0,1]$ and then $f(0)=f(1)$ then what is the difference
16:39
there is no difference
I think $[0,1]$ actually fails something, but I can't recall it for now
by the universal property of quotients
There is no difference, I just find $I^n$ nicer to think about than $S^n$
@LeakyNun yeap the formalism is what gets me alot. i can talk the talk but i cant walk the walk most times.
@ManolisLyviakis aha, the formalism
16:40
And it's natural to define a loop has having domain $[0,1]$ if you already defined paths first
contrary to popular belief, topology isn't geometry done without lengths, angles, and proofs
certainly without lengths and angles, but more certainly with proofs
that's not really a popular belief, it's a joke
i never really got involved with geometry but topology just sucked me in
ima take a course on differential now
and certainly i want to redo complex analysis with all of the algebraic tool i have now from topology
differential geometry
differential geomtry is harder than algebraic geometry for me
it must be alot more fun doing complex analysis with loops and paths
i have not touched algebraic geometry :P seems like an outland or something
16:44
25 mins ago, by MatheinBoulomenos
at least there are surjective continuous functions $[0,1] \to S^2$
and then just join $f(0)$ and $f(1)$ with a straight line
and then you obtain a loop covering $S^2$
@LeakyNun so you agree that the problem on the proof was that there are space filling curves?
@ManolisLyviakis yes
or you can find something else?
ok thanks
do you remember that we did the latter part of the proof before?
I gave you a thousand functions and composed them together
16:48
should we repeat it?
to prove that a function no surjective to the sphere is homotopic to a constant
sure !
So we know that $\Bbb R^2$ is isomorphic to $S^2 \setminus \{pt\}$, i.e. $f:\Bbb R^2 \cong S^2 \setminus \{pt\}$
isomorphism means that it has an inverse $g : S^2 \setminus \{pt\} \cong \Bbb R^2$ such that $f \circ g = \operatorname{id}_{S^2 \setminus \{pt\}}$ and $g \circ f = \operatorname{id}_{\Bbb R^2}$
16:51
now we have a loop $X : S^1 \to S^2$, where $pt$ is not in its image
which means they have the same homotoy type
I haven't finished
do you know what a pullback is?
nope ive heard it before
but i cant recall
16:55
basically, in this context (there is a more general context), the pullback of the diagram $X \overset f \longrightarrow Z \overset g \longleftarrow Y$ is $P = \{(x,y) \mid f(x) = g(y)\}$ with projection maps $P \to X$ and $P \to Y$ sending $(x,y)$ to $x$ and $y$ respectively
so in particular, this is a pullback diagram: $$\begin{array}{c} P & \overset {\pi_x} \longrightarrow & X \\ \downarrow {\small \pi_y} && \downarrow f \\ Y & \overset g \longrightarrow & Z \end{array}$$
So pullback is a space
Here: $$\begin{array}{c} P & \overset {\pi_x} \longrightarrow & S^1 \\ \downarrow {\small \pi_y} && \downarrow X \\ \Bbb R^2 & \overset f \longrightarrow & S^2 \end{array}$$
brb
convince yourself that $P = \Bbb R^2$ for now (I can't reply as often) @ManolisLyviakis
sorry, that isn't $f$
that is $f$ precomposed with the inclusion $S^2 \setminus \{pt\} \to S^2$
im trying !! :D
17:12
0
Q: Solving differential-difference equation using Fourier series

Lozansky Determine a solution with period $2$ of the differential-difference equation $$y'(t)+y(t-1) = \cos^2 \pi t$$ Attempted solution The r.h. side can be rewritten as $\cos^2 \pi t = \dfrac{1}{2} + \dfrac{1}{4}e^{2\pi i t} + \dfrac{1}{4}e^{-2\pi i t}$. With period $2P = 2$ we determine the funda...

Does anyone see what I did wrong?
@ManolisLyviakis oops, it isn't $\Bbb R^2$ I think
@Lozansky from what I know of a similar kind of problem in physics, your period 2 solution should end up having $y(t+1)=-y(t)$.
now, $P = \{(t,y) \in S^1 \times \Bbb R^2 \mid X(t) = \iota(f(y))\}$, where $\iota : S^2 \setminus \{pt\} \to S^2$ is the inclusion map
so indeed $\iota(x) = x$ for all $x \in S^2 \setminus \{pt\}$
for every $t \in S^1$, we can find exactly one $y$ such that $X(t) = \iota(f(y))$
because $X$ is a function from $S^1$ to $S^2$
so indeed $P \approx S^1$
Hey chat, is it an issue when a grade school teacher marks 5*3=5+5+5 wrong as opposed to 3+3+3+3+3 on the basis of problem comprehension?
@ManolisLyviakis are you with me?
that's just a convoluted way of saying "from the loop $X : S^1 \to S^2$ construct the loop $X' : S^1 \to \Bbb R^2$"
by "removing" $pt$ from the codomain of $X$
17:27
Though maybe that's only true when it's $y''(t)$ not $y'(t)$...hrm
@ManolisLyviakis now, $X'' : S^1 \times I \to \Bbb R^2$ is the homotopy from $X'$ to the constant map
yeah, I think I'm being too influenced by a different problem.
@Semiclassical I don't think that is true
I don't either at this point.
@ManolisLyviakis and then $\iota \circ f \circ X'' : S^1 \times I \to S^2$ will give you the homotopy from $X$ to the constant map
17:30
@LeakyNun what you're doing is a really convoluted way to compose with the inverse of the isomorphism $S^2\setminus\{pt\}\approx \Bbb R^2$
@Semiclassical I'm a bit skeptical if it's possible to construct such a solution. $\cos^2 \pi t$ has period $1$, not $2$
hmm
if there's an escape clause here, I think it'll be because the ansatz you're using isn't generic enough
i.e. you should be including more stuff than just $\{e^{i n \pi t}\}$
But it's not obvious to me why anything else is permissible.
Eh, this is an exercise in Fourier series with other periods than $2 \pi$
@MatheinBoulomenos but that's the proper way :P
17:34
why?
It could be noted that $cos^{2} \pi t$ is even, so we are only looking for a solution with $\cos$ terms
Morning
@KasmirKhaan you alive? or did algebra finally do you in?
@MatheinBoulomenos pull the loop $X:S^1 \to S^2$ back along the inclusion $\Bbb R^2 \to S^2$
I understand what you're doing. I don't quite understand why you're doing it
because how else would you construct the loop $X' : S^1 \to \Bbb R^2$?
17:39
we're already assuming that the image is contained in $S^1\setminus \{pt\}$
so we compose with the stereographic projection $S^1\setminus \{pt\} \to \Bbb R^2$
@Lozansky this depends a bit on the wording of the problem, of course. I mean, your solution is 2-periodic in that $y(t+2)=y(t)$. The problem is that it has least period 1, not least period 2.
@MatheinBoulomenos how do you restrict the codomain?
you mean how do we get continuity if we restrict the codomain? That follows from the definition/universal property of the subspace topology. If $X \subset Y$ has the subset topology, then a function $Z\to X$ is continuous iff the composition with the inclusion $Z\to Y$ is
@Semiclassical Yeah, I think it makes sense that the solution is $1$-period since the R.H is. But could we just as well have looked for a $1$-period solution and found the same one?
hmm
part of what makes me worry here is that the inhomogeneous term in your equation is built out of the same functions as your solution is
and that can make things a bit hinky.
17:48
@MatheinBoulomenos so you're saying that $X$ is a subspace of $Y$ iff there is a split monomorphism $X \to Y$ in the category of topological spaces?
I'm not sure what to conclude, though.
Hopefully someone else can shed some light.
@LeakyNun that's not what I'm saying
then what is it?
@Lozansky looking at some stuff quickly, one suggestion I'm seeing that makes a lot of sense is to work via Laplace transforms. I might play around with that and see what I get.
@MatheinBoulomenos is it a functor of some sort?
17:53
If you want to state it categorically, subspaces are the subsets such that the inclusion $X \to Y$ is a regular monomorphism
@Semiclassical Yeah I was thinking that too. Only I'm "supposed" to work it out using Fourier series
Right.
A Laplace transform solution might at least shed light on what might be being missed, though
but if you actually want to prove that maps are continuous what I wrote above is more useful
@MatheinBoulomenos prove
@Semiclassical Not sure what you are going to do with the $y(t-1)$ term though
17:57
Yeah.
The inlcusion $X \to Y$ is the equalizer of the projection $Y \to Y/X$ and a constant map which sends everything in $Y$ to the class of $X$ in $Y/X$
You could write it as $y(t-1)\theta(t-1)$, of course your solution is only valid for $t>1$ then
Yeah. The problem with the Laplace transform idea is that it doesn't play so well with periodicity
but the stuff about monomorphisms is not important for restricting the codomain
you can just use the definition of subspace topology?
so maybe it's better to bite the bullet and look at the Fourier transform
18:00
Yeah
I haven't really proved that my solution is indeed correct
I just tested quickly for $t=1$ which worked
Which definition of the Fourier transform do you know, if any? There's a few different conventions for it.
This chapter is way before the transform
The definition we use is $\hat{f}(\omega) = \int_R f(t) e^{-i\omega t}$
18:07
Oh
Yeah
And $f \in L^{1}(\textbf{R})$
18:28
Hey y'all, what is the technical name for the open $n$-dimensional disk? i.e. the set
$D = \{(x_{1},x_{2},\cdots,x_{n}) : \sqrt{(x_{1} - x_{1_{0}})^{2} + (x_{2} - x_{2_{0}})^{2} + \cdots + (x_{n} - x_{n_{0}})^{2}} < d \}$
Where $(x_{1},x_{2},\cdots,x_{n})$ are all variables and $d > 0$
I get that one dimension would be a line, two corresponds to a "real" disk, and three would be a "sphere", but I'm not sure what comes at $n$ dimensions
Ball?
A big suprise : $f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?
@Lozansky if that's the technical name, then yeah!
@CookieToast Isn't it just an n-sphere?
@CryinShame I have no idea. Your guess is as good as mine. To be fair, the wikipedia page for a "Ball" like Lozansky suggested seems to describe what I'm talking about
In mathematics, a ball is the space bounded by a sphere. It may be a closed ball (including the boundary points that constitute the sphere) or an open ball (excluding them). These concepts are defined not only in three-dimensional Euclidean space but also for lower and higher dimensions, and for metric spaces in general. A ball or hyperball in n dimensions is called an n-ball and is bounded by an (n − 1)-sphere. Thus, for example, a ball in the Euclidean plane is the same thing as a disk, the area bounded by a circle. In Euclidean 3-space, a ball is taken to be the volume bounded by a 2-dimensional...
I'm guessing then that both terms are synonymous with eachother
18:39
@Cookie no mine is wrong. The sphere is at equality. I think you got it will the ball.
Ah, because a sphere only describes the points bounding the region, and not the interior?
Yeah I imagine that would have been the motivation.
Either way, more terminology learned = good stuff! Thank you
No no problem, you ended up teaching me so thanks in return!
Would it be weird to refer to a region as an $n$-ball? For example, the open region bounded by the sphere would be a $3$-ball?
18:44
does anyone know how I can create a bigger version of $\Sigma$
Ah, wikipedia seems to use that terminology, so I'm guessing its acceptable
@adeek This question on the TeX SE has some solutions, but they're for actual LaTeX files and wouldn't be usable in this chat
35
Q: making a big summation sign

Preludehow can I make a big summation sign ? ‎\begin{align}‎ ‎\cos x‎ ‎=‎ ‎‎\sum\limits‎_‎{n=0}^{‎\‎‎infty‎} ‎\frac{(ix)^2n‎}{(2n)!}‎‎ \end{align}‎ this is the code which I'm using for summation

I've had the same question for years my dude :)
Thanks @CookieToast
just what I need
It would be cool if there was some sort of extension for this chat where you could include your own desired LaTeX packages
yeah that would be awesome
hi
I've got a question about undefined expressions
18:59
@parvin just ask it
there is this phrase : $ 1/a - 1/b $ , and limit for $b$ is infinit
as far as I know the subtraction is allowed when is equal to $ 1/a + 0 $
Let $(X,\tau)$ be a topological space, and $A \subseteq X$. Then, the interior of $A$ is $\displaystyle \bigcup \{x \in \tau \mid x \subseteq A\}$
@parvin yes
well I guess I got my answer already :D
thnks any way
@Lozansky I went and did the solution via Fourier transforms, and I get the same solution as you. So this, coupled with LutzL's additional commentary, pretty much settle things.
Hi guys. there is this Lemma which says you can delete finitely many terms from a convergent series and it will still converge. This does not imply it will converge to the same limit though right?
19:07
@philmcole Right
to see an example, just consider a series where all but finitely many terms are 0
@TobiasKildetoft Thanks!
what up peeps
Hey @BalarkaSen
I am just writing couple of things for my master thesis
would you like to take a look at it It is like first 7 pages
Not particularly, @Adeek
I have a bunch of things to do
oh okay
19:17
@Semiclassical Very nice! But how did LutzL conclude $g_+(t) = A + B\cos(2 \pi t) + C\sin(2 \pi t)$? Just an ansatz for $\cos^2(\pi t)$?
Not sure.
0
Q: If the solution is too simple, it must be incorrect?

user193319 Prove that to each $\epsilon > 0$ there exists a $\delta > 0$ such that $\displaystyle \int_E |f| d \mu < \epsilon$ whenever $\mu (E) < \delta$. I found this question asked on MSE here, here, and here, but some of the suggestions seem relatively complicated (at least compared to what I did)....

Well as $\cos^2{\pi t} = \dfrac{1}{2}(1+\cos(2\pi t))$ I guess that makes a lot of sense
Although I think one should first find solutions to the homogenous equation before making such an ansatz
19:57
you know @Semiclassical one great thing one can do is download many phd thesis in your topic of study and read them
just to get idea of what is happening
Hey how i show that the only two clopen sets of $\Bbb R^n $ are $\Bbb R^n $ and $\emptyset $
\emptyset: $\emptyset$
or \varnothing $\varnothing$
Thanks ^^
i was able to show $R^n $ and $\emptyset$ were clopen but i dont know how to show they are the only two. i assumed that there existed anther set that was clopen and then it compliment must also be clopen
but beyond that im not sure, i noticed a proof for $ \Bbb R $ involved using sup but im not sure what that means in $R^n$ ?
20:20
"Find the Fourier series of the odd function of period $2$ that is described by $f(t) =t(1-t)$ for $0 \leq t \leq 1$"
How can that function possibly be odd?
Well, if it were odd, what would it need to look like for $-1\leq t<0$?
$t(t-1)$
Oh yeah that works
Thought it would mess up the periodicity
in that case you'd have f(-1/2)=(-1/2)(-1/2-1)=3/4 and f(1/2)=1/4
so that's not quite right
Ok $t(t+1)$
check: f(-1/2)=(-1/2)(1/2)=1/4 = -f(1/2)
yeah, you're good
20:25
Whats it mean for a sequence in $\Bbb R^2 $ to be bounded?
more simply, you want $f(t)=t(1-|t|)$ for $-1\leq t\leq 1$
can you imagine extending that 2-periodically?
Yes, graphically I can
that result is what you want to construct the Fourier series for.
Yeah ok, we only get sine terms
20:29
$b_n = 2\int_{0}^{1} t(1-t) \sin n\pi t dt = \dfrac{4(1-(-1)^n)}{n^3 \pi^3}$
seems plausible, I don't know off the top of my head.
note that this restricts you further to odd $n$
I.e. $f(t) = \dfrac{8}{\pi ^3} \sum_{k=0}^{\infty} \dfrac{\sin(2k+1) \pi t}{(2k+1)^3}$
right
probably you want to write that as sin[()\pi t]
but anyways.
A nice little check here is that you should have $f(1/2)=1/4$. But in that case the sum becomes $\sum_{k=0}^\infty (-1)^k/(2k+1)^3$
And then you're doing Riemann zeta-type stuff
That is a standard sum?
Oh yeah gotcha
Standard enough, I think
though, n^3...hrm
usually when you get n^3 you don't get sums which can be done in closed form
but Mathematica verifies that the sum works out, so I guess I'm wrong
20:36
@Semiclassical $f(1/2) = 1/4 = 8/ \pi^3 \sum (-1)^k/(2k+1)^3 \to \sum (-1)^k/(2k+1)^3 = \pi ^3/ 32$
yeah, Mathematica verifies that
So that's a good sign.
Yeah that's a relief
I'm gonna read about Gibbs phenomenon now ^^
have fun
21:02
In case any of you know ML...
0
Q: Find the MSE of a true response and its predicted value using OLS estimation

ClarinetistFrom Theodoridis' Machine Learning, exercise 3.26. Consider, once more, the same regression as that of Problem 3.8, but with $\boldsymbol\Sigma_{\boldsymbol\eta} = \mathbf{I}_N$. For context, this is the regression model $$y_n = \boldsymbol\theta^{T}\mathbf{x}_n + \eta_n\text{, } \qquad n...

(Machine Learning)
21:38
Fun fact: This is not a geodesic tiling
(i.e., those white lines are not geodesics of $\Bbb H^2$; they meet the boundary of the disk at ~80 degrees)
local teenager EVISCERATES Escher's tesselation of hyperbolic space
more at 11
It's actually Coxeter
He showed there's no such tiling which alternates between squares and triangles
Kinda cool
local acclaimed geometer EVISCERATES Escher's tessellation of hyperbolic space
more at 11
Oh. Maybe it's a group theory fact. I guess the argument goes like, find the symmetry group of such a possible configuration if there is such a thing, and prove that it's not a subgroup of the isometry group of $\Bbb H^2$, i.e., $\text{SL}_2(\Bbb R)$
@Fargle looool
Let's see. There's a 4fold symmetry coming from the squares by hyperbolic rotation around the center and a 3fold symmetry from the triangles...
$a^4 = b^3 = 1$... p sure it should be a quotient of $\Bbb Z/4 * \Bbb Z/3$
Im too lazy to figure out what it actually is
rip

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