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12:00 AM
oh
I think i Get it
He means "hypermatrices"
Not $3 \times 3$
 
@HenryT.Horton He means tensors, as anon says.
 
Yes I realized as soon as I posted the link
 
@BenjaLim Hola!
 
Hello ugly
 
@HenryT.Horton "- Quick call him ugly! - You're chuby!"
 
12:11 AM
I'm ugly and chubby
 
@HenryT.Horton Good. So what?
 
Just trying to fit in
 
@HenryT.Horton Fits like it was made for it.
 
Benjalim is gone... All the unanswered questions...
 
12:23 AM
Gone?
 
No longer a part of this chat...
 
@HenryT.Horton Yo, yo.
 
As of this moment or as of forever?
 
I almost finish an exercise
 
As of this moment.
He didn't even say bye...
What exercise, my bald fellow math.SE chatter?
 
12:30 AM
Bald?
 
@HenryT.Horton TO prove $\operatorname{int}(A)$, $\operatorname{int}(X\setminus A)$ and $\partial A$ are mutually disjoint, and then prove $\operatorname{int}(A)\cup\operatorname{int}(X\setminus A)\cup\partial A=X$,
 
@HenryT.Horton yes?
 
@BenjaLim We thought you were gone.
 
Benjalim?
 
12:57 AM
@HenryT.Horton perhaps his connection dropped.
 
We lost him...
 
@HenryT.Horton He might be playing with the transcript and only popping in to say something.
@PeterTamaroff Are you defining $\partial A$ as $\overline{A}\cap\overline{X\setminus A}$?
 
I think I have an answer to a question, but I'm scared
 
@HenryT.Horton Can I ask you a quick question?
 
The problem with answering Makoto's questions is that they get punted off the main page realllly fast.
 
1:06 AM
Are you a ghost?
 
If you have say the function $f : \Bbb{R}^n - \{0\} \to \Bbb{R}^n$ given by $f(x) = x/||x||$
I want to say the quotient of two continuous functions is continuous
but the problem is that the guy on the numerator is a function from $\Bbb{R}^n$ to $\Bbb{R}^n$
 
@BenjaLim You do not want that formula at $0$.
 
while the denominator is a function from $\Bbb{R}^n$ to $\Bbb{R}$
@DylanMoreland the zero vector is excluded a priori
 
Then you should delete it from the domain.
At any rate, one way to go is to say that a map into the product is continuous iff the components are.
 
@DylanMoreland I want to use that to give me a deformation retract of $R^n - \{0\}$ onto $S^{n-1}$
 
1:08 AM
Now everything is into $\mathbb R$ and you're very happy.
Indeed it does.
 
yeah
@HenryT.Horton The individual coordiantes
are functions from $\Bbb{R}^n$ to $\Bbb{R}$
and are continuous
the numerator is the coordinate function which is continuous
 
Yes... quotients of continuous functions where the denominator is nonzero on the domain
 
and the denominator clearly a continuous functions
so it's good
 
So good.
 
@HenryT.Horton That's my deformation retract!!
 
1:11 AM
Not yet
 
@HenryT.Horton ?
 
You want a homotopy from the identity function to your $x/\|x\|$
 
Well actually to $i \circ x/||x||$
@HenryT.Horton but that inclusion map is annoying
it does nothing
 
You do nothing
 
huhuhuh???
 
1:14 AM
Fight me
 
I want $i \circ x/||x|| \simeq id_{\Bbb{R}^n - \{0\}}$
 
Why the inclusion map, what is it including
 
$i : S^{n-1} \rightarrow \Bbb{R}^{n} - \{0\}$
That's the definition of a deformation retract that I have @HenryT.Horton
 
Then throw away your book
Or wherever you got it from
 
It's rotman
btw
@HenryT.Horton Hmmm the problem is now figuring out the homotopy
 
1:19 AM
You want a continuous map $F: (\Bbb R^n \setminus \{0\}) \times [0,1] \longrightarrow \Bbb R^n \setminus \{0\}$ such that $F(x,0) = x$ and $F(x,1) \in S^{n-1}$ for all $x \in \Bbb R^n \setminus \{0\}$ and $F(a,1) = a$ for all $a \in S^{n-1}$
The homotopy is easy
 
yes
I am trying to figure that out homie @HenryT.Horton
 
So in particular you want $F$ to be a homotopy from the identity to $x/\|x\|$
Do you want to know the answer or will you flail around like a fish a little more
2
 
I want to figure it out by myself
@HenryT.Horton Take $$F(x,t) = \frac{(1-t)x}{||x||} + tx$$
that does it
 
yes
 
1:22 AM
It's backwards
 
@HenryT.Horton doesn't matter
 
-5 points
 
@HenryT.Horton Then do $$(1-t)x + \frac{tx}{||x||}$$
@HenryT.Horton You like to troll very much .......
 
-1 point for getting in my face
 
hahahahahahhahahahaha
 
1:24 AM
It's not trolling, it's making sure your solutions are precisely correct
 
well but whether the homotopy is from a to b or b to a
to say that two continuous functions from $X$ to $Y$ are homotopic is an equivalence relation
 
Well I told you the direction I wanted to h*motopy to go, and I'm the boss
 
@HenryT.Horton Ok Al capone
 
These aren't paths
But homotopy of maps is an equivalence relation
 
yes
 
1:26 AM
You didn't show me your proof of that yet though.
 
proof of??
Suppose $f,g$ are maps from $X \to Y$ and are homotopic via $H(x,t) : X \times I \rightarrow Y$
Then $H(x,1-t)$ is a homotopy between $g$ and $f$ and so $g \simeq f$
 
Make that $f_t$ into $F_t$, don't use $f$ for two things
Hopefully that $H$ stands for the Horton Homotopy
 
@HenryT.Horton better?
@HenryT.Horton No.
@HenryT.Horton I don't flail around like a fish because I'm not magikarp
 
So homotopy of maps is a symmetric relation
 
yes
 
1:29 AM
One day you'll be a Gyarados. If you can handle me as your trainer.
 
@HenryT.Horton No you can't even catch me with a masterball :D
 
I'd just leave you in the computer box for the whole game...
 
Tossing would be bad for the environment?
 
@HenryT.Horton Ok I'm off to show that $S^n$ is a retract of $D^{n+1} - \{0\}$
@HenryT.Horton wait wouldn't my map that I used before also prove this fact?
 
Yeah ***
 
1:33 AM
WTF....................................
 
Look who came crawling back
 
No I was like wait a minute it's exactly the same map
@HenryT.Horton wait a minute
wait
I can use my map above to show that $S^{\infty}$ deformation retracts onto $S^0$?
 
If $x \in D^{n+1}$, then $x/\|x\| \in D^{n+1}$ and since your homotopy was a convex combination of $x$ and $x/\|x\|$ and $D^{n+1}$ is convex, the homotopy still works
No
$S^{\infty}$ is contractible
 
yeah but wouldn't that show it?
 
$S^0$ isn't contractible
It has the homotopy type of TWO points
Not one
 
1:36 AM
Oh I forgot that if two spaces have the same homotopy type
then if one is contractible the other is
 
The first issue is: what would $\|x\|$ mean for $x \in S^\infty$
 
yeah....
 
And what do you mean with $S^\infty$
Unit sphere in a separable countably infinite Hilbert space?
And even if it did work, how would you get a deformation retract from $S^\infty$ onto $S^0$?
 
yes I got mixed up with my definitions
 
You're high
 
1:39 AM
I got confused between two spaces having the same homotopy type and two maps being homotopic
 
Well if $A$ is a deformation retract of $X$, then $X$ and $A$ are homotopy equivalent
 
yeah
that's obvious from the definition
and from the definition you see why you need the inclusion map
 
Are you still rambling on about that inclusion map!?
 
yes
Because recall to say that two spaces are homotopic
you need to have a map $f : X \rightarrow A$ and $g: A \rightarrow X$ such that their compositions are homotopic to the identities
And if you take $g$ to be the inclusion map then if $A$ is a deformation retract of $X$ you also get that they are homotopy equivalent
@HenryT.Horton If $X$ is contractible then $X$ has the homotopy type of a point yes?
 
1:43 AM
converse?
 
@HenryT.Horton $S^{\infty}$ is contractible and hence has the homotopy type of a point
 
$X$ is contractible $\iff$ $X$ has the homotopy type of a point
 
So why can't $S^{0}$ that is a point be a deformation retract of $S^{\infty}$? @HenryT.Horton
 
hi
 
1:45 AM
@RajeshD hey
 
whatz up, topology?
 
$S^0 = \{-1\} \cup \{1\}$...
$S^0$ is the set of all points in $\Bbb R$ that are at a distance of $1$ from $0$
 
what??????????
Oh crap
@HenryT.Horton Ok but what if I show that $S^{\infty}$ deformation retracts onto say $(1,0,0,\ldots)$
 
lol
 
@HenryT.Horton I thought S^0 was just a point oh god facepalm.....
 
1:48 AM
You FOOL
 
@HenryT.Horton Ok I show it deformation retracts onto a point
that would show it has the homotopy type of a point yes? @HenryT.Horton
 
Yes... what did you show deformation retracts onto a point
 
the problem now is I am not really sure what $S^{\infty}$ is, hatcher doesn't really define it in chapter 0.... @HenryT.Horton
he just says it is $\cup_{n \geq 0} S^n$
 
Is this from Hatcher? Your problems?
You know how to define the topology on that union, right?
 
2:02 AM
Well dawg I'm just gonna go now...
I have to tutor in the morning and then I need to figure out the stuff I am going to lecture about for my reading course
Bye... forever
 
2:30 AM
@HenryT.Horton Why forever?
 
2:44 AM
@HenryT.Horton No actually I don't know the topology on the union
 
@robjohn Yes. But I'm watching Transformers DOTM now!
 
Why is any irrep of $G$ (finite group) a direct summand of some tensor power of a given faithful rep?
Qiaochu asked for compact groups, and the answers on MO are thus over my head.
 
3:01 AM
Hi, everybody.
Is it a new result?
 
what?
 
yo = hi
 
Can someone help me understand a matter in Robert israel's answer here?
He says "Since (the function $f$ under consideration) is analytic and not constant, it has finitely many zeros in a compact set. So any plane has only finitely many intersections with (the graph of $f$)."
I do not understand what is being concluded in the first sentence. Does he mean that $f$ has only finitely many zeroes overall, or just that given some compact set $C$ it has only finitely many zeroes in $C$?
 
 
2 hours later…
5:21 AM
Ah, the room slowly repopulates...
 
Hi @robjohn
 
@skullpatrol howdy
 
@robjohn How's it going?
 
@skullpatrol pretty well. On vacation.
 
5:33 AM
@skullpatrol we're in the mountains, but the paradigm is easily shifted.
 
Don't you love vacations? @robjohn
 
@skullpatrol they are very nice.
 
user19161
5:49 AM
@skullpatrol Interestingly that desktop background is really one.
 
@JasperLoy Indeed :-D
 
user19161
@skullpatrol If that tip were so narrow he would have fallen off by now.
 
@JasperLoy Agreed.
@robjohn Enjoy your vacation.
 
@skullpatrol thanks! I'm sure we will.
 
user19161
@robjohn Wait, why are you in chat when on vacation? :-)
 
5:57 AM
@JasperLoy He's on vacation, not in exile!
 
user19161
@skullpatrol One can be in exile and in chat bro!
 
@JasperLoy Addiction is a terrible thing. :-D
 
user19161
@robjohn Yeah, I have tried unsuccessfully to quit chat/SE.
 
Tell me about it.
A.K.A. Math Addict fallen off the wagon.
 
i was successful
 
6:05 AM
@Eugene Congratulations.
 
well goodnight
 
6:26 AM
my brother has been listening to the same rap song for over an hour. I want to strangle him.
2
 
@anon Rap really does make people violent huh
 
@anon Listening to the same 3 minute song for over an hour is a form of brain conditioning/washing, in my opinion. How old is he?
 
17
 
Challenge him to write the lyrics on paper.
Without the song playing of course.
 
6:43 AM
Good morning
 
6:59 AM
@anon Hey
Can I ask you something quick?
 
go ahead
 
We say that $f_1(x),f_2(x)$ are linearly dependent if there exist $a,b$ such that for all $x$, $af_1(x) + bf_2(x) = 0$
now the negation of this
is that the functions are linearly independent
 
mmhmm
 
should that be for all $a,b$ such that FOR ALL $x$, we have $af_1(x) + bf_2(x) \neq 0$? @anon
No I think the FORALL should be there exist....
 
it means for any $a,b$, the combo $af_1+bf_2$ is not the zero map. this doesn't mean it never takes on a zero value, as nonzero maps can have zeros. nawatImean?
 
7:01 AM
yes.
@anon yes.
 
7:14 AM
Hi.
 
$(\exists a,b)(\forall x) af_1(x)+bf_2(x)=0$
To get the negation you simply change quantifiers and negate the claim. It's pretty mechanical.
$(\forall a,b)(\exists x) af_1(x)+bf_2(x)\ne 0$
 
What books, links would you recommend for the approaching of the multisums? I'm also interested in some new research on this topic.
 
By multisums you mean something link $\sum_{i=1}^n \sum_{j=1}^i a_{ij}$?
 
Yes. With 2 or more sums.
 
I believe that exercises on this topic can be found in many introductory texts on discrete mathematics (for finite sums) or calculus (for both finite in infinite sums).
 
7:31 AM
OK. Thanks.
 
7:43 AM
@Chris'sister I have some doubts about possibilities of doing research on multiple sums. (It is pretty vague description of what the research should be about.)
But maybe summability methods for double sequences are close enough to this and it would be possible to make some research in this area. (Not sure to which extent one can expect some useful results.) I'll list a few papers I am aware of. Most of them just randomly caught my attention in the past and I make a note; it's more a sample than a suggestion where study of this are could start.
Boos, Leiger, Zeller: Consistency theory for SM-methods, Acta Mathematica Hungarica, 1997; doi: 10.1007/BF02907056
Zeltser: On Conservative Matrix Methods for Double Sequence Spaces, Acta Mathematica Hungarica, doi: 10.1023/A:1015636905885
Patterson: Analogues of some fundamental theorems of summability theory, link
J. D. Hill: Almost-convergent double sequences; Tohoku Mathematical Journal, Volume 17, Number 2 (1965), 105-116. link
Perhaps you'll get some other suggestions from other users.
And perhaps you should clarify what you mean under being interested in new research on this topic. Do you plan to write some kind of project/essay/thesis and you want choose an area which is interesting for you?
 
8:02 AM
@anon How's your urge to strangulate going? May I suggest ear plugs?
 
nah, he went to bed a bit ago
 
He's probably having "rapper dreams."
 
8:19 AM
Okay nevermind same old Makoto.
 
hm?
 
@Dylan Perhaps your comment here could be an answer - so that the question is not left unanswered.
 
@DylanMoreland Does that guy never stop :(
 
I can't tell if he's asking a question of his own invention or some kind of exercise from a textbook.
 
Maybe he is trying to write his own textbook, and just testing the problems here first?
Using us as proof-readers?
 
8:40 AM
Ignorance is bliss. I know almost nothing about commutative algebra, so I was pretty much unaffected by Makoto Kato's questions.
According to Wikipedia Makoto is a unisex Japanese given name. So Makoto might be she.
 
It might be a pseudonym. It's very transparently the word for "truth".
 
9:07 AM
@JonasTeuwen Good morning - how are things today?
 
9:20 AM
I feel quite uh wasted. 8-). Good morning, how are you?
 
Oh pretty good here, thanks - really enjoying one of yesterdays books
 
:-). Nice!
 
@JonasTeuwen Sorry to hear "wasted" :(
 
Oh. I don't know. Maybe just tired.
 
9:33 AM
"Tired and emotional" is a British euphemism for "drunk", or so I'm told... :p
 
I must go - back later
 
See you! :-).
@ZhenLin I am not emotional :P. And I didn't drink any alcohol in many weeks!
 
hey guys?
i need help on improving my answer here - math.stackexchange.com/questions/175385/…
 
10:13 AM
@jokerdino Save your efforts because the "powers that be" will [close] that question soon.
 
aww okay
 
Thanks. Watching.
 
11:13 AM
Hi there!
 
japp
 
Perhaps introducing new (makoto-kato) tag would solve some of the problems. ;-)
 
@draks Oh hi, I have not seen you here before...
 
No problem Can anybody help me with a wiki page?
 
11:23 AM
@MartinSleziak Not unlike that "pony" tag in meta... :)
@draks which?
 
@J.M. I don't understand why this guy makes you so angry.
 
:5516805 It was quite apparent, no worry...
@MarkDominus I'm not, actually. I just find the damn thing funny.
 
I guess I misread.
 
@draks Yes, it's from the asymptotic series for the exponential integral.
@MarkDominus s'okay. If he were driving me nuts, I'd have been writing a thread-ful of grievances in meta... ;)
 
11:26 AM
They say: In mathematics, the divergent series ,lalala
and then: first considered by Euler, who applied resummation methods to assign a finite value to the series.
What now?
Does converge to the value given at the end or diverge?
 
@draks Yeah, Borel is quite convenient.
 
Let me amend that: I don't understand why this guy makes so many people so angry.
I must use the site differently because I have barely noticed him.
 
Isn't the box the right place to cool him down?
 
@draks The term I am accustomed to is that the exponential integral is an antilimit for that series. The series itself is divergent (being an asymptotic series), but 1. it can be made sensible, and 2. it can be analytically continued.
@MarkDominus Have you seen Martin's survey in meta?
 
Just saw, thanks.
 
11:29 AM
@draks Moderators are certainly aware of him. The decision is up to them, I'd say.
 
@draks It's apparently at least gotten rid of one problem among many.
 
@draks Hockey penalty box? Or chain gang sweatbox?
 
@J.M. are you talking the wiki page now?
 
@draks I'm looking at it. It's the same as saying "$\zeta(-1)$ is the antilimit for $1+2+3+\cdots$"
@MarkDominus I think a literal sand box. "Now, sweetie, go play in the sandbox while Daddy's working, okay?"
 
@J.m. sorry I'm not familiar with an antilimit.
that's not even on the wiki...
 
11:34 AM
@draks As I said: the series is not convergent in itself, but a certain value can be associated to it via, say, analytic continuation of an equivalent form. That value is the antilimit.
 
@J.m. aha
and because there is an antilimit
one can interchange summation and integration?
 
Actually, one could say that we are heavily abusing the notion of $=$ for antilimits...
@draks No, it's more a formal manipulation at that point.
 
Honestly, The whole page sounds to me like doing one error after another
and at the end all cancel
at least kind of.
 
@draks Have a look at this answer and see if that helps to understand any more about divergent series.
 
To go back to your original example, the actual development would be to repeatedly integrate the exponential integral by parts.
 
11:37 AM
@robjohn thanks, might take me a while
@J.M. ah that's better now
 
@draks, if you have Mathematica on you, try executing Series[ExpIntegralEi[x], {x, Infinity, 15}]. Notice anything?
 
@J.M. no I don't. Usually I ask Wolfram...
 
It should work in Wolfram Alpha as well, I think.
 
it does
but you meant -EI(-x), right?
 
@J.M. It does
 
11:41 AM
@J.M. errh x=-1 I mean
 
@draks try this link.
 
@robjohn thanks
@robjohn doesn't work, but don't mind
I already got it copy-pasted
 
@draks $\mathrm{Ei}(x)=-E_1(-x)$.
...to use the two different kinds of exponential integrals.
So, formally substituting $x=1$ into that asymptotic series nets you something "equal" to $e \mathrm{Ei}(1)$
 
@J.M. one is enough, I once got confused by all these
 
@draks Sorry, sometimes I forget that special functions can be overwhelming...
 
11:47 AM
and wiki pages can be full of confusing nomenclature...
 
@draks Sorry, the link got messed up by my browser, I think this should work.
 
@robjohn this one works, thanks
 
Again, having the divergent series and the integral on both sides of the "$=$" is somewhat of an abuse of the notion of "$=$", but there we are.
 
Hello folks!
 
11:49 AM
ok thank you guys
 
@KannappanSampath Hello.
 
Just so much tired!
 
How is this site ever going to put a rest to the 48 ÷ 2(9+3) question? I mean it just keeps popping up over and over again.
 
@skullpatrol propose a 48/2(9+3) tag!
 
@draks Sounds like a good idea :-D
 
11:54 AM
@draks Now that question is definitely annoying.
(...and why I loved Gerry's answer.)
@Jonas: it's a bit offputting when even the supposed experts have no clue, no?
 
Yep.
And some even say "just wait".
 

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