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17:00
@ManolisLyviakis I think so
So say i have the $R^2$ without the point zero
it isn't
?
i was gonna say
go on
consider 2 functions
one above the point zero
and one below
obviously i cant pass the point
but cant i say the one which is above zero is homotopic to a certain constant (point)
then that point to another
go around
and when im finaly below the zero
then that point is homotopic to the function below zero?
so the two of them are homotopic after all
why i cant do that?
17:04
because I have no idea what you mean by a function above zero
f(x)=(cos(πx),sin(πx)) be the half
circle
g(x) be the other half
domain?????
the line in $R^2$
don't miss out every other detail and expect us to be able to get you
17:06
which line?
live?
conside the 2 halfs of the uni circle as 2 functions
unit*
domain???
in $R^2$
$R$
is the domain i guess
17:07
what?
@ManolisLyviakis ...
is it R or is it [0,2pi]?
can you think about your question before you ask it?
i did
Consider the Unit Circle around zero In the punctured $R^2$ without the point zero. And consider the 2 functions the upper half of the circle and the lower half .
17:10
what function?
17:20
why i cant deform f to a point and move it around ?
ask munkres for the domain xD
path has implicit domain [0,1]
you hardly used the word "path"
you can't deform it to a point because you can't expand a point back to a line
i just said consider the 2 halves of the unit circle :p
aww
thanks
"as functions"
i cant deform the point back to a line but i can make the other line a point
no, a half of the unit circle isn't just any old function
@ManolisLyviakis yes you can
17:26
and i cant deform these two points to each other?
you can
but then whats the mistake?
deform $f$ to point $h$ to apoint and the 2 points to each other with a line
there is no mistake
im just starting understanding how homotopy works
17:28
it says that $f$ and $h$ are not homotopic
there exists no path-homotopy between paths $f$ and $h$
ohh
yes
haha
omfg
it says they would be homotopic to $R^2$
i was talking about the punctured $R^2$
which means i cant do what i said in $R^2$
but i dont know why
what is "path homotopy"?
you mean that if
i do my way it wont satisfy the point wise relation?
hmmm thanks yes
guys can someone please help with a seeming loophole I ran into?
2
Q: Getting 2 different answers when solving a differential equation in 2 different methods

RickI was trying to evaluate $ {\int_0^{\infty}\frac{\cos (x)}{1+x^2}}\text{d}x$ by Feynman's trick It has been solved here, but I have a specific question at one of the steps while doing it. (Not a duplicate) So I defined $I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$ and arrived at $I(t...

17:33
@LeakyNun you could have said that earlier you made this really hard xDDD
I'm not seeing it
of course there is a path-homotopy between f and h in X
I'm saying there is a path-homotopy between f and h in X
I don't agree with the book
ah
there isn't one that fix both end-points
if you deform them to points and go around then propably you wont satisfy the requirement that thehomotopy must satisfy
the ending and starting point
the homotopy needs to preserve end-points
17:36
yes exactly !!
you can't deform any one of them to a point
when you asked me about the definition i saw it too xD
you should check the definitions before asking :P
so they are homotopic but they aint path-homotopic
wow i feel stupid XDDD
right
17:37
"right" hahahaha
I mean, right they aren't path-homotopic
i know i was joking :D cause of the timing
can i ask one more thing?
@LeakyNun ?
if i prove 2 functions are homotopic to a space Y
are they also homotopic to a homeomorphic space of Y?
I think so
17:44
example : Let $f:X-->s^n$ not a surjective map then how do i proceed to prove it is homotopic to .. say the constant function $c:X---> S^n$
i was thinking moving the problem to $R^n$ through the stereographic pojection
and then using the straight line homotopy
but i dont know how
pick a point not mapped
then stereographic project it with respect to that point
then you have R^n, done
not mapped by f
right
project the whole sphere without that point in $R^n$
17:47
then what do i say i mean the math i cant write them down
f goes to the sphere not to $ R^n$ but i have projected its image in $R^n$
it is actually a homeomorphism
S^n\{pt} is homeomorphic to R^n
yes so? how do i say that now my $f$ is $ :X-> R^n$
compose the maps lol
17:50
f composition stereographic projection isnt thtat a different map than initial f?
i want to prove homotopy for my initial f
i know it might seem trivial to u but im missing the small steps i of coursed thought of composition
$f : X \to S^n$
$g : \Bbb R^n \cong S^n \setminus \{pt\}$
$h : S^n \setminus \{pt\} \hookrightarrow S^n$
$i : X \to S^n \setminus \{pt\}$
let $pt$ be the point not mapped to
then $f$ descends to the map $i$
such that $f = i \circ h$
now compose $i$ with $g^{-1}$
and you obtain $g^{-1} \circ i : X \to \Bbb R^n$
where it is homotopic to a point with homotopy $\varphi : X \times I \to \Bbb R^n$
i is the identity?
where $\varphi(x,0) = g^{-1}(i(x))$ and $\varphi(x,1) = O$ where $O$ is the origin
no, $i : X \to S^n \setminus \{pt\}$
$i$ and $h$ what do these do?
now compose $\varphi$ with $g$ to obtain a homotopy $\varphi' : X \times I \to S^n \setminus \{pt\}$
$h$ is the inclusion map that doesn't depend on $f$
$i$ is essentially the map $f$
but we don't need that extra point in the codomain
17:56
oh ok im starting to get it
now compose $\varphi'$ with $h$ to obtain a homotopy $\varphi'' : X \times I \to S^n$ and you're done
wow you found those maps really fast :P
so you went to R^n
and then back again
right?
found a homotopy to $R^n$ and then a homotopy back to $S^n$
yeah i wouldnt have thought of that i got as far as intuitivevly understanding that i can project them and find i homotopy in $R^n$ but i knew it wasnt for my initial $f$ didnt thought of going back to $S^n$
$$X \overset f \longrightarrow S^n$$
$$\begin{array}{c} X \\ \small i \downarrow & \searrow \small f \\ S^n \setminus \{pt\} & \underset h \hookrightarrow S^n \\ \small {g^{-1}} \downarrow \small \sim \\ \Bbb R^n \end{array}$$
$$X \times I \longrightarrow \Bbb R^n \overset g {\underset \sim \longrightarrow} S^n \setminus \{pt\} \overset h \hookrightarrow S^n$$
18:19
wow awsome
thanks
 
1 hour later…
19:26
If a question asks for 3 proper divisors, is 41,43, and 47 (for example) valid?
Or do we have to include 1? Making it 4 proper divisors?
proper means non trivial
most cases
So if a number has only 41,43,and 47 for it's divisors, that's valid right? we don't include 1?
no
it means except itself
4
has 1,2
oh ok so the max is 47*43.
Ok thank you @ManolisLyviakis
19:41
nice try tho
20:02
How do I find the number of integer triangles, none of whose sides exceed 4?
@Fargle (removed)
combinatorics :P
I am trying to use the triangle inequality.
$a+b\ge c$
for the one side you can have 1,2,3, or 4 as a side
But question imposes a constraint, $a,b,c \in \{1,2,3,4\}$
20:04
4 options for each side
$ 4^3$ ?
unless u mean a sum or something
that's not the right answer.
and I am sure thats not the way to solve it.
u can have 0?
no
then why not $4^3$
ohh
i got it
u cant have 1,1,3
then for the first side
u have 4 options
for the second also 4
but for the third
u can have all the possible number who are greater than a-b
and less than 4
so you can have 4-1 but then you can choose only 3 or 4
so you can count those possibilities
What is going on
20:21
Hey @Balarka
Do you have a moment for a quick doubt on Aldehydes and Ketones?
Jesus christ
He actually left the chat
True story.
Also, hi @Alessandro
Welcome back :P
:P
How's it going?
20:25
I should be studying more functional analysis for the exam in January but I'm not, so great!
I guess that'll be my new year's resolution...
Functional analysis is deadly
Functional analysis is p gr8 tho
Some things are cool but other are rather ugly (Ascoli-Arzelà...)
Arzela Ascoli isn't ugly
I also really dislike the style of the professor but maybe that's just me
20:33
it's both easy to prove and useful
What does it say
I know 0 functional
Arzelà-Ascoli looks completely unmotivated to me, we haven't seen any application or corollary
It characterizes the (relatively) compact sets in $\mathcal C(X,Y)$ where $X,Y$ are metric spaces, $X$ is compact, and you use the sup norm on the set of continuous functions $X\to Y$
Oh, this sounds interesting.
it is like immensely useful
So what are the compact sets of that?
20:35
I have applied it at least a hundred times
bounded + equicontinuity are the conditions
Pointwise compact+equicontinuous
the version you quoted is fancy
90% of the time $Y = \mathbb{R}$
I think the one Alessandro quoted is an iff thing
At my uni, once an algebraic geometer taught functional analysis. He used sheaves to define distributions and worked with exact sequences of Banach spaces ...
20:38
Alessandro: do you know about compact operators?
I will soon @Dami :P
@Daminark as opposed to what
i dont know any version of AA that isnt an iff
Hm, pointwise compact implies bounded if $Y=\Bbb R$?
Lolol. Well, the quick version is this, an operator $T:X\to Y$ is compact if the image of a bounded set has compact closure. Now, use A-A to prove that the adjoints of compact operators are compact
idk what pointwise compact means
20:41
So a subset $K \subset C(X, Y)$ is equicontinuous means that any sequence of functions in $K$ is an equicontinuous family $f_n : X \to Y$?
Or what
It means that fixed $\varepsilon$ the same $\delta$ works for all the functions in $K$
That seems to be what I said.
@EricSilva $\{f(x)|f\in K\}$ is compact in $Y$ for all $x$
this is what bounded means to me
Ah, no, sorry.
Equicontinuous families give rise to equicontinuous subsets of C(X, Y)
This is a generalization where the index is over... K
Makes sense
20:46
@BalarkaSen do you know if I need to know chapter 1 of Hartshrone to read chapter II ?
I don't. I have never read Chapter II, only I
can I jump right away to chapter II ?
I see
isnt chapter 2 the bad boi with the scary rep
It's schemes
it doesn't look scary
20:47
It is though.
It's nontrivial to understand.
Doesn't bounded mean $||f||<M$ for all $f\in K$?
i would call this uniformly bounded
@ManolisLyviakis i get 14 pairs.
Right, what this means is that every $f \in K$ is bounded by something.
Also Eric: I didn't know that the A-A in Rudin was iff
20:48
Note that every Banach space is isometrically isomorphic to a closed subspace of the continuous functions on a compact space, so this makes Arzelà-Ascoli more widely applicable
the AA in rudin is that you can extract a convergent subseq iff youre bounded + equicontiunuous @Daminark
im p sure
This seems like something which should immediately be very useful to me
But maybe that's because I have thought a lot about functions spaces for a few weeks
Anyway my "fancy" version is actually the same version then @Eric
yeah their proof is the same
but i barely ever see it applied in the fancy form
brb
20:53
@Narcissusjewel Check out "Mirror Reaper" by Bell Witch.
@Eric Is there an Arzela-Ascoli if I want to look at $C^r(M, N)$ (where $M$ and $N$ are smooth manifolds)?

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