« first day (2202 days earlier)      last day (3117 days later) » 
00:00 - 17:0017:00 - 00:00

Huy
Huy
17:00
I am
what is holiday season
there'll be enough time for holidays when I retire
So, I want to finally sit down and do some topology
@Huy What are you doing?
@BalarkaSen Why aren't you doing that
Huy
Huy
@Krijn: Farb&Margalit as always, just going through something I discussed with Balarka earlier
@Krijn i'm lazy
Hey there, i'm wanting to create one single function where : if x < 0 -> f(x) = 1/x, and else f(x) = x. How could i do that ?
17:03
You just created one.
lol
@PearlSek are you aiming to somehow make those pieces continuous?
yes, but the thing is that i need it to keep this in one function
what would you exactly mean by 'one' function?
I don't know what you mean by one function. What you did is a perfectly fine definition: it's called defining a function piecewise.
17:05
It is however, not continuous on $\mathbb R$
Both the pieces are continuous, in fact smooth. You cannot make it continuous at $0$ because $1/x$ blows to infinity.
Huy
Huy
no, to minus infinity
perhaps you could restrict the domain where it fits $\frac{1}{x}$ to some point before $0$
like i would like to keep it on a single f(x) = statement
@Huy How do you know those are different?
17:06
@Huy Yes, thanks for correcting.
Huy
Huy
$$f(x) = \begin{cases} \frac{1}{x} & x < 0\\x & x \geq 0 \end{cases}$$
@Krijn: I checked thoroughly
@BalarkaSen is here $X = M$ or $X = T$?
(the first one)
$M$, sorry.
@PearlSek if you want a single definition, here's one, although it's a bit funny if this was the answer that would be accepted as correct-
$f(x) = \frac{|x|-x}{2x}\cdot \left(-\frac{1}{x}\right) + \frac{|x|+x}{2x}\cdot x
*$f(x) = \frac{|x|-x}{2x}\cdot \left(-\frac{1}{x}\right) + \frac{|x|+x}{2x}\cdot x$
the only problem is at $x= 0$
@SoumyoB You can also just edit messages
right I forgot about that
17:12
I like you solution though!
If it was f(x) = x if x >= 0 (instead of x > 0), easier would have been $x^{x/|x|}$. With the convention 0/0 = 1, it's fine.
You could simplify it using a $\operatorname{sgn}$ function
@BalarkaSen That's horrendous and beautiful at the same time
Huy
Huy
@BalarkaSen sorry if this is just being pedantic, but here, $F$ should go to $S^1 \times (0,1)$, right? if yes, isn't a deformation retraction usually going to a subspace?
or do you actually mean $S^1 \times \{0,1\}$
@BalarkaSen that was so much more original than my version of Langrange interpolation!! (I don't mean it sarcastically)
17:17
@Huy Yes and no. If $A \subset X$, a deformation retract of $X$ to $A$ is a homotopy $F : X \times I \to X$ between the identity map so that $F(-, 1)$ is a retract (image of $F(-, 1)$ sits in $A$).
Huy
Huy
yes
So yes, I do actually mean $S^1 \times \{0, 1\}$. That's what the retract goes to.
Huy
Huy
ah, ok
(the boundary of $S^1 \times [0, 1]$, aka)
$F$ is simply a deformation retraction of $S^1 \times I$ minus center circle to $\partial{S^1 \times I} = S^1 \times \{0, 1\}$, in short. You can write down such a def. ret. by just working with the longitudes ($x \times [0, 1]$) of the cylinder.
Huy
Huy
17:19
and $F(x,1)$ is just any retraction of $S^1 \times \{0,1\}$?
Well, I have a specific retraction in mind (obtained by pushing the two components in the complement of center circle longitudinally towards the boundary) but sure.
"retraction of S^1 x {0, 1}" is an odd parsing though.
Huy
Huy
yeah
Did you mean "retraction to"?
Huy
Huy
yes
OK, cool. So yes.
Huy
Huy
17:21
$\partial S^1 \times I = S^1\times \{0,1\}$???
$\partial(S^1 \times I)$, sorry.
Huy
Huy
ok, thanks
I almost had a stroke
Huy
Huy
I already drew pictures trying to figure out how that is true
17:23
typos are always a pain in the neck
@Semiclassical Hi!
Hi, @SemiC.
@Huy how did you put that boundary notation in latex?
Huy
Huy
partial
$\partial $
oh I see thanks
17:26
you can always right-click mathjax to see what the tex was
@Semiclassical can you detail a bit more on how to do that?
>right-click >Show Math As >TeX Commands
Huy
Huy
@BalarkaSen: sorry, I'm still a bit confused by your notation. just for the start, $F$ is the homotopy between the identity and the retraction $r: S^1 \times (0,1) - S^1 \times \{ \tfrac{1}{2}\} \to S^1 \times \{0,1\}$, right?
17:42
Yes.
Huy
Huy
@BalarkaSen: and how do you define $F(x,0)$? you say it's the inclusion, but the boundary of $S^1 \times (0,1)$ isn't a subspace, technically?
ah, wait
You're right, I lied. I was working with $S^1 \times [0, 1]$, not $S^1 \times (0, 1)$. But this is not a big issue, in the sense that there are arbitrarily small tubular neighborhoods so that you can always choose a tubular neighborhood so small that it's closure is topologically $S^1 \times [0, 1]$.
So feel free to replace $S^1 \times (0, 1)$ by $S^1 \times [0, 1]$ throughout.
Huy
Huy
I might just, to start with
(Otherwise replace $S^1 \times \{0, 1\}$ by $S^1 \times (0, 1/3] \cup S^1 \times (2/3, 1]$)
@Krijn roger that, thanks
18:06
What should i do when i encounter idiots like this guy http://math.stackexchange.com/questions/1890888/trigonometry-problem-of-higher-secondary

just look the comments made by the questioner
I don't know what you should do, but you certainly you should NOT be offensive towards the OP
but he is brainless
Doesn't justify your offense.
and his offense ?
I agree the question is ill-formatted and the OP is not actively doing anything to make it better: I would have voted to close if I were to vote for anything. But I do not think he said anything offensive: I would have flagged your message too if I had anything to flag.
18:10
"Good for you , now put your penis in your mouth as a reward" :)
Huy
Huy
@BalarkaSen: ok, so you first show that there is a def. ret. from $S^1 \times [0,1]$ minus center circle to $S^1 \times \{0,1\} \subset S^1 \times [0,1]$ and then apply that to get an analogous def. ret. in a surface
isnt this offensive
I don't see that message. Perhaps mods deleted it?
@Huy Yup
Huy
Huy
@BalarkaSen: and before you wrote up that argument, you said "deformation retract doesn't induce a deformation retract of the complement always"
so retracting the boundary to the chain is the deformation retract, and what we've done here is the deformation retract of the complement
Yes.
But this is only particular to this case, with a chain and it's tubular nbhd
There are cases where it fails.
Huy
Huy
18:13
why does that now prove it? does a deformation contract leave connected components invariant?
ah, is it continuous?
Oh, yes, it does.
Huy
Huy
ah, I see
I totally forgot
Homotopy equivvalent spaces always have the same # of connected components.
Huy
Huy
I see
damn someone has been trolling the stack exchange forum I see
Huy
Huy
18:14
@BalarkaSen: but doesn't that require me to show that cutting along the boundary doesn't disconnect the surface, in the first place?
so are the moderators going to lift the ban hammer? I'm curious to see what happens
Huy
Huy
if the scc did cut it, we have 2 components
the case described by @A---B
Huy
Huy
so the def. ret. shouldn't change that?
"bro just leave it , run off your business" he still has some slightly offensive comments I see
hasn't yet deleted them
how curious
18:16
@Huy Careful now. We showed that the complement of the chain deformation retracts to the complement of the tubular neighborhood of the chain, note. Now complement of the tubular neighborhood of the chain is connected because it's one of the two connected components of the complement of boundary of the chain.
That implies complement of the chain is connected, is what I am trying to say.
We're not looking at both the components of the complement of the boundary circle. Just one component: the exterior one - complement of the other component (aka the tubular nbhd of the chain)
(and even then, we're pretending we proved this for a chain: we just proved this for a chain of length 1 aka an scc)
Huy
Huy
yes, that last part's clear ^^
give me a second, I don't think I understand it yet but don't know what question I should ask
Sure. Let me know if I need to clarify or summarize what we proved and what we didn't.
@BalarkaSen happy independence day btw :)
Huy
Huy
@BalarkaSen: the retraction $F(x,1): S^1 \times [0,1] - S^1 \times \{1/2\} \to S^1 \times \{0,1\}$ "pushes" points towards the boundary of $S^1 \times [0,1]$. wouldn't the other direction be the desirable one, here? (because we're moving the boundary of the tub nbhd towards the circle?)
@SoumyoB Yep, we're ""free"" today.
@Huy We're deformation retracting $M - C$ to $M - T$, not the other way around.
$C$ is contained in $T$: hence $M - C$ contains $M - T$.
The way to do that is to "push stuff away from $C$"
aka, locally, push complement of center circle in $S^1 \times [0, 1]$ away towards boundary
Huy
Huy
18:32
ok
hi @BalarkaSen
Huy
Huy
yes, that I understood
let me see if I still don't understand how the statement then follows (for the circle)
ok
Hi @Adeek
16 days until grad school starts so excited.
Huy
Huy
18:33
ok
and you're saying because $M-T$ is connected, $M-C$ must be by continuity
I have been sitting at home for many days studying and solving problems yeah but I would prefer going to university.
I am reading a cool math history book by hawking it is very nice.
Huy
Huy
and $H(x,1)$ is the deformation retract
@Huy yes, htpy eq preserves # of path-connected components (continuity imo is a bit of understatement there, but sure)
@Huy $H$ is the deformation retract
Huy
Huy
right
$H(-, 1)$ is the retract.
Huy
Huy
18:37
that "retract map" $H(x,1)$ goes from where to where?
$M - C$ to $M - T$.
Huy
Huy
ok, ok
I think I understood the argument
Huy
Huy
thanks !
sure, sure.
Huy
Huy
18:39
so to summarise: "this follows from basic topology which I've never bothered to learn from Hatcher yet"
Huy
Huy
next holidays I'll start with Hatcher, for real this time
seems to be a lot of useful stuff in there :>
good, good
yup, there are
Huy
Huy
@BalarkaSen: in the case of even curves in a chain, is it even possible that cutting along the tub nbhd doesn't separate the surface?
what do you mean by "cutting along tubular neighborhood"? complement of tubular neighborhood of an even length chain is connected because it's a component of the complement of the boundary circle of the tubular neighborhood (along with the other component, namely the tubular neighborhood).
Huy
Huy
18:49
along the boundary
(that's unrelated to the proof)
@Huy So you're asking if it's possible that complement of boundary of tubular nbhd of an even length chain is connected?
Huy
Huy
yes
even length chain
Obviously not, but I'd be a bit hard pushed to write down a proof. I mean, the boundary of the tubular nbhd obviously bounds the tubular nbhd itself :)
So you're essentially asking if the tubular neighborhood is diffeomorphic to all of the complement of the chain. That can hardly ever happen.
I don't see what's particular about "even length" here though.
user227867
I got 200 points for a simple answer, lol.
Huy
Huy
19:05
it's not difficult to find a single curve that doesn't disconnect a surface
but you're talking about boundary of tubular neighborhood of a chain, not the chain itself.
Huy
Huy
oh
right
@BalarkaSen Ignore this, by the way.
Huy
Huy
that's a new kind of "note to self"
19:17
@JasperLoy Should not impress you after all this years.
Huy
Huy
@BalarkaSen: I've gone through the proof again a few times. I'm convinced. thanks for your help again :)
why do you think I'll need to apply the deformation retractions simultaneously ?
@Huy no problem
@Huy you won't, not really. you need to do that, along with a lot of bump function trickeries, to make your deformation smooth, but of course that's not required if you want to prove that # of connected components are the same
just a topological deformation retraction is enough
Huy
Huy
19:33
yes, that's what I thought
good thinking
Huy
Huy
but enough thinking for today
"think, think. practice thinking."
 
1 hour later…
r9m
r9m
20:53
@user1618033 Book!! be like ..
21:07
@r9m o/
21:27
hello all. Was wondering if anyone knows what way would be most effective for finding the value of X in AX=B, where you know the value of A and B?
Huy
Huy
depends on what A X and B are
sorry my apologies, yes that would be helpful. Both a and b are matrices, a is a 3x3, while b is a 1x3 matrix
Huy
Huy
that's not possible
oh. well that would explain why I couldn't find a solution.
and I made a mistake in typing. B isn't a 1x3, it's a 3x1
Huy
Huy
then it is possible
21:38
I guess that brings me back to my original question, being what would the most effective process be for determining the value of x in this instance where a and b values are known. I'm not looking for a solution, more verifying if the processes that I've been using are correct.
Huy
Huy
Gauss elimination would be a simple way
Better then the inverse method that I was told earlier. I'll give Gauss elimination a look see. Thanks.
Huy
Huy
inverse method only works if A actually is invertible
which is not the case very often
Ya that's what prompted my question. A lot of the questions that I have that method runs into a brick wall.
 
1 hour later…
23:09
(Message 1/3) All users, I add here these calculations, if some user want to do this exercise. I don't know if it is useful this recipe. I start with a trigonometric identity, type for example in Google Books, the words: Murty, Problems in Analytic Number Theory (Springer). Then you can see in page 41, the Exercise 3.2.10.
@user243301 No, don't do this.
(Message 2/3) Do the specialization with $\theta=\pi s$ and combine it with the functional equation for the Riemann zeta function. Take the absolute value of previous, and combine it with Proposition 1 of Nazardonyavi and Yakubovich On an inequality for the Riemann zeta-function in the critical strip, this article from arXiV is arxiv.org/pdf/1206.1801.pdf .
Why are you doing this?
Okey @PedroTamaroff, if I disturb to you my apologizes.
Well, not me.
23:10
Hi @PedroTamaroff
Then my apologizes all usrs.
It just seems that if we all left arbitrary calculations in chat for no reason whatsoever, chat will soon collapse with such things.
Okey, I understand, and I am agree with you.
@BalarkaSen Hello.
I learnt a thing or two about harmonic functions lately.
Complex-analytically oriented things, that is to say.
23:14
Cool potatoes and beans.
How about you?
What have you been learning, of late?
00:00 - 17:0017:00 - 00:00

« first day (2202 days earlier)      last day (3117 days later) »