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Balarka Sen
Balarka Sen
6:00 AM
By "algebraic-topology", most MO users would understand "chromatic homotopy theoretic topology quantum ring spectra".
 
Anthony
Anthony
@BalarkaSen Are you any good with functional analysis?
 
Balarka Sen
Balarka Sen
No, @Anthony.
 
Anthony
Anthony
Ah
 
Balarka Sen
Balarka Sen
@Rememberme If you mean Schubert's calculus, no I don't.
I have heard about it, but that doesn't imply I know about it.
 
Remember me
Remember me
Thats a part of Combinatorics which has just been found and answer to the questions in it involves a lot of algebraic topology and algebraic geometry!!
Combinatorics with topology!!!! Great combination
 
Balarka Sen
Balarka Sen
6:04 AM
Suggest you study combinatorics/algebraic topology/algebraic geometry individually first before getting so excited about all of them :P
 
Remember me
Remember me
No my brother introduced me to this........
 
Disciple of Barney
Disciple of Barney
I don't suggest that
 
Kaj Hansen
Kaj Hansen
Are you around @BalarkaSen ?
 
Balarka Sen
Balarka Sen
Yeah.
 
Kaj Hansen
Kaj Hansen
I have a really silly question, or at least it feels silly
 
Balarka Sen
Balarka Sen
6:11 AM
okay.
 
Kaj Hansen
Kaj Hansen
As part of a problem I'm working on, I need to show $\frac{|z|}{|z^4 + 10z^2 + 9|} \rightarrow 0$ as $|z| \rightarrow \infty$.
 
Balarka Sen
Balarka Sen
try dividing numerator and denominator by $1/|z|$
and then make $z$ huge. see what happens.
 
Kaj Hansen
Kaj Hansen
Divide? Or multiply?
 
Balarka Sen
Balarka Sen
multiply, sorry.
what d'you get in the denominator?
 
Kaj Hansen
Kaj Hansen
The denominator becomes $\frac{|z^4 + 10z^2 + 9|}{|z|}$
 
Balarka Sen
Balarka Sen
6:15 AM
now chuck the $1/z$ inside the absolute value
 
Kaj Hansen
Kaj Hansen
$|z^3 + 10z + 9/z|$
 
Balarka Sen
Balarka Sen
that $9/z$ is small, i don't care about it. do you?
 
Kaj Hansen
Kaj Hansen
Well, I want to be very careful here since we are in complex world. Certainly that'd be true for real numbers.
 
Balarka Sen
Balarka Sen
oh, you are doing this for complex numbers?
 
Kaj Hansen
Kaj Hansen
Indeed
 
Julian Rachman
Julian Rachman
6:17 AM
Hey. Do any of you know any good math poems? I need one to analyze for class
 
Kaj Hansen
Kaj Hansen
That limit is trivial for $\mathbb{R}$
 
Julian Rachman
Julian Rachman
Hey @Kaj and @Balarka
 
Kaj Hansen
Kaj Hansen
Hey @JulianRachman
 
Balarka Sen
Balarka Sen
fair
 
Kaj Hansen
Kaj Hansen
 
Disciple of Barney
Disciple of Barney
6:18 AM
@JulianRachman here
 
Kaj Hansen
Kaj Hansen
Just having problems proving it.
 
robjohn
robjohn
@KajHansen If $|z|\to\infty$, $\frac1{z}\to0$ even in $\mathbb{C}$.
 
Balarka Sen
Balarka Sen
you have to use some absolute value ineqs, in that case.
 
Kaj Hansen
Kaj Hansen
Certainly @robjohn. But it's inside an absolute value with some other stuff, so I cannot consider it stand-alone.
 
Julian Rachman
Julian Rachman
@DiscipleofBarney Thanks. And I have been tinkering with Beamer and it is cool.
 
Balarka Sen
Balarka Sen
6:19 AM
@Kaj why not?
$9/z$ tends to something zero.
the rest you have inside absolute value and is huge.
inverting gives something small.
 
robjohn
robjohn
@KajHansen separate it with the triangle inequality. Then it is standing alone and still small
 
Disciple of Barney
Disciple of Barney
@JulianRachman Glad to help. What were you planning on making into a poster (what is it about)
 
Balarka Sen
Balarka Sen
Hi @PedroTamaroff
 
Kaj Hansen
Kaj Hansen
I tried that @robjohn. Doing so makes the denominator *larger*, which makes the whole fraction smaller:

$\frac{|z|}{|z^4+10z^2+9|} \geq \frac{|z|}{|z^4| + |10z^2| + 9}$ no?
So the RHS certainly tends to zero, but says little about the LHS.
 
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen Hey.
 
Julian Rachman
Julian Rachman
6:24 AM
@DiscipleofBarney I was not quite sure yet but I can across something about Alzheimers and the retrival of memory (there is a long story that goes with how I got to what I had come across). So I want to see if I can develop a mathematical model to help unfold the mysteries of memory lose due to Alzheimers.
 
Balarka Sen
Balarka Sen
Haven't seen you for a while in here.
What have you been thinking about lately, then?
 
Julian Rachman
Julian Rachman
It was just something I thought of and my present at like a local science fair maybe. Along with other research. @DiscipleofBarney But it is only a dream now.......
 
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen Are you asking me?
 
Balarka Sen
Balarka Sen
yeah
 
Pedro Tamaroff
Pedro Tamaroff
Well, that's a very broad question.
I've been thinking about a lot of things.
 
Balarka Sen
Balarka Sen
6:26 AM
How about the "in mathematics" condition imposed?
 
Pedro Tamaroff
Pedro Tamaroff
It's still broad.
 
robjohn
robjohn
@KajHansen $$\frac{|z|}{|z^4+10z^2+9|}\le\frac1{|z|^3}\frac1{1-10/|z|^2-9/|z|^4}$$
Which is small for $|z|$ large enough. Is there some other estimate you need?
 
Disciple of Barney
Disciple of Barney
Sounds cool. At the JMM this year I saw a presentation on this group using persistant homology to study blood-flow, I think, in the brain and I guess it ended up far more accurate than previous models (like orders of magnitude). I don't remember all the details, I think they were comparing different ages and their blood-flow. Maybe there is something to that and Alzheimers? (I am taking a wild guess as your project got me thinking of that talk) @JulianRachman
 
Pedro Tamaroff
Pedro Tamaroff
At the moment I'm trying to decide what chapter to read in a book, @BalarkaSen.
 
usukidoll
usukidoll
matlab driving me coo coo insane ughhhh
 
Balarka Sen
Balarka Sen
6:29 AM
Which book, @Pedro?
 
Pedro Tamaroff
Pedro Tamaroff
Rotman's book on homological algebra.
 
Balarka Sen
Balarka Sen
ah.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff Go with the next one :)
 
Balarka Sen
Balarka Sen
I don't know anything about homological algebra. Would love to know, though.
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Hey there. =)
 
usukidoll
usukidoll
6:31 AM
@DiscipleofBarney my graph gets the point right but it looks crooked and shifted
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff Ohh, and hi
 
Disciple of Barney
Disciple of Barney
@usukidoll I don't know anything about matlab
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft How is it going?
@BalarkaSen Well, you'll get there.
 
Balarka Sen
Balarka Sen
hopefully.
 
usukidoll
usukidoll
oh x.x
 
Pedro Tamaroff
Pedro Tamaroff
6:32 AM
Just don't put the carriage before the horse.
 
usukidoll
usukidoll
oh sorry I am confusing you with another user
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff Good. Hopefully I should soon be getting around to putting some papers together
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Sounds exciting.
I am writing, but on a different level.
 
Balarka Sen
Balarka Sen
@PedroTamaroff i won't, i promise. i am going to learn some multivariable calculus first. :P
how are you going with Hatcher, btw?
 
Pedro Tamaroff
Pedro Tamaroff
So as not to be too self promotional, I'll just tell you to look into my MSE profile for a link, @TobiasKildetoft.
@BalarkaSen I never started reading it.
I just read a section I needed.
 
Balarka Sen
Balarka Sen
6:34 AM
where are you studying homology from, then?
 
Kaj Hansen
Kaj Hansen
Excellent @robjohn. Nice use of the reverse triangle inequality. Much appreciated.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff Neat. I am thinking of writing a second blog post here on the math SE blog
 
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen Weibel and Rotman. Hatcher is a book on algebraic topology, not really about homological algebra.
 
Remember me
Remember me
The path of love is never smooth
But mine's continuous for you
You're the upper bound in the chains of my heart
You're my Axiom of Choice, you know it's true

But lately our relation's not so well-defined
And I just can't function without you
I'll prove my proposition and I'm sure you'll find
We're a finite simple group of order two

I'm losing my identity
I'm getting tensor every day
And without loss of generality
I will assume that you feel the same way

Since every time I see you, you just quotient out
 
Balarka Sen
Balarka Sen
@PedroTamaroff Oh, so you are studying homological algebra, not singular homology. I see.
 
Pedro Tamaroff
Pedro Tamaroff
6:36 AM
@BalarkaSen Yes.
@TobiasKildetoft I saw some mean comments on a post.
Not sure it was you.
But people were being asses, as usual. =/
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff I commented about the newest one in the blog chatroom, but only about the typos (I find no problem with the exposition)
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Sure. Comments were not about typos. =P
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff Yeah, I also found the comments on the post itself tasteless. Constructive criticism can be fine, but this was not constructive at all.
 
Balarka Sen
Balarka Sen
hi, @Mike, sorry for being silly yesterday. was sleepy.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff so what sorts of homological algebra are you studying now?
 
Balarka Sen
Balarka Sen
6:46 AM
we had found a homeomorphism type for SP^n(C^*), not SP^n(S^1)
 
Mike Miller
Mike Miller
No worries, @Balarka; sorry for being impatient.
Yes.
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft So far in the course we saw the basics, Ext, Tor (which we balanced), a bit of bicomplexes, the total tensor and hom complexes, and the Künneth formula.
I'm reading Weibel and Rotman on my own, though, plus the course's exercises.
Here's the page of the course.
Professor leaves lots of exercises which is nice.
 
Disciple of Barney
Disciple of Barney
@JulianRachman here is the page (it is the jmm slides). Pretty sure it is this one.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff I forgot the generality those books work in. Are these for commutative rings, or generally. Or in full generality in abelian categories?
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Mostly for module categories, but one also looks into abelian categories. I haven't read too much, though. Rotman's I've read the first three chapters, and Weibel I'm halfway through the second.
 
Tobias Kildetoft
Tobias Kildetoft
6:54 AM
@PedroTamaroff Well, abelian categories are close enough to being module categories anyway, via some embedding theorems
or using some double centralizer property if the category is nice enough
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Right, yes. I want to read about Freyd Mitchell in detail, but it seems most books only skim through it.
@TobiasKildetoft I don't know what that is.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff I have not actually ever looked that kuch at the embedding theorems
the double centralizer is that if you have a projective generator $P$, then the category is equivalent to modules over the endomorphism algebra of $P$ (under some extra conditions I always forget)
(and sometimes one can take something other than a projective generator)
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Ah, I know this one... =)
Something something Morita, something something.
@TobiasKildetoft We actually peeked into that in Mariano's course.
Given an algebra $A$, we decomposed it as sum of projectives stemming from primitive orthogonal idempotents (so the projectives are indecomposable) and took one copy of each isoclass of projectives to get a basic algebra with the same module category, using that you say.
@TobiasKildetoft The notes are in english.
 
Julian Rachman
Julian Rachman
@DiscipleofBarney That's great! Thanks. @Rememberme Thanks. I will. Who is it by?
 
Remember me
Remember me
I just found it on math overflow and thought it might help you @JulianRachman
But the poem is amazing
 
Julian Rachman
Julian Rachman
7:02 AM
Ok! THanks!
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff cool. A third alternative is to work with members, which allows one to do most diagram chases directly in the abelian category instead of using an embedding.
 
Julian Rachman
Julian Rachman
Ok. Gtg everyone. See everyone tomorrow morning (or whatever your timezone may be) :D
 
Tobias Kildetoft
Tobias Kildetoft
though they do take a bit to get used to
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft What are members?
Never heard about those.
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff a member of an object $X$ is an arrow $a: Y\to X$ for some object $Y$
 
Pedro Tamaroff
Pedro Tamaroff
7:04 AM
@TobiasKildetoft OK.
 
Tobias Kildetoft
Tobias Kildetoft
and two members $a: Y\to X$ and $b: Y'\to X$ are equal if there is an object $Z$ and arrows $c: Z\to Y$ and $d: Z\to Y'$ such that the obvious diagram commutes
these then behave in many ways like elements of sets do in module categories, when it comes to questions of monomorphisms, epimorphisms and exactness
 
Chris's sis
Chris's sis
@robjohn did you manage to get some progress on that integral?
 
Tobias Kildetoft
Tobias Kildetoft
They are introduced nicely in McLane's book on categories for the working mathematician
 
robjohn
robjohn
@Chris'ssis I worked a bit and then got taken away. I have not finished it.
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Ah. I have yet to learn seriously about categories. I know tiny bits here and there, but eventually I'll have to learn about them properly.
 
Chris's sis
Chris's sis
7:07 AM
@robjohn OK. Have you found a promising way?
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff I found McLane's book a good read, but it does take quite a bit of background to understand all the examples and really see the point of it all
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft @TobiasKildetoft I'll have that in mind. I have to sleep now! As I told @KarlKronenfeld, feel free to meanly criticize my blog posts if you have the time to. >=)
 
Tobias Kildetoft
Tobias Kildetoft
@PedroTamaroff I felt I needed to get a better grip on them now that I am working on 2-representations of 2-categories
 
Pedro Tamaroff
Pedro Tamaroff
@TobiasKildetoft Hahaha, good luck with that.
Now I'll have nightmares!
runs away in fear of 2-representations
 
Remember me
Remember me
Hi @Incurrence
 
Incurrence
Incurrence
7:12 AM
@Rememberme Hey
 
Anthony
Anthony
7:33 AM
 
Chris's sis
Chris's sis
7:43 AM
An unsimplified component of the answer $$1/24 (96 Catalan \[Pi] + \[Pi]^2 Log[1/2 - I/2] - \[Pi]^2 Log[
1/2 + I/2] + 48 Log[1/2 - I/2]^2 Log[1/2 + I/2] -
48 Log[1/2 - I/2] Log[1/2 + I/2]^2 - 25 \[Pi]^2 Log[1 - I] +
24 Log[1/2 - I/2]^2 Log[1 - I] -
48 Log[1/2 - I/2] Log[1/2 + I/2] Log[1 - I] +
24 Log[1/2 + I/2]^2 Log[1 - I] - 23 \[Pi]^2 Log[1 + I] -
24 Log[1/2 - I/2]^2 Log[1 + I] +
48 Log[1/2 - I/2] Log[1/2 + I/2] Log[1 + I] -
24 Log[1/2 + I/2]^2 Log[1 + I] -
48 Log[1/2 - I/2] PolyLog[2, 1/2 - I/2] -
 
Remember me
Remember me
What on earth is this?@Chris'ssis
 
Chris's sis
Chris's sis
@Rememberme Some calculations.
 
Remember me
Remember me
These hugeee?
 
Chris's sis
Chris's sis
@Rememberme It's just a component of an answer.
 
Remember me
Remember me
Just a component!!!!!!!!
 
robjohn
robjohn
8:13 AM
@Anthony sorry...
 
N3buchadnezzar
N3buchadnezzar
8:53 AM
Gah, I hate being beaten when writing a longer answer. Gah math.stackexchange.com/questions/1249478/…
 
Chris's sis
Chris's sis
Some upvotes needed here (not for me, but for the question)
0
Q: $\int_0^{\pi/2} (2 x+\pi \sin (x)-2 \pi ) \sec (x) \log (\cos (x)) \, dx$

Chris's sisHow would you like to tackle the following integral $$\int_0^{\pi/2} (2 x+\pi \sin (x)-2 \pi ) \sec (x) \log (\cos (x)) \, dx \ ?$$ I thought of some complex analysis, but still not a nice way to go.

calculus real-analysis integration definite-integrals
 
Chris's sis
Chris's sis
9:19 AM
$$\huge{\text{CHRISTTTTTTTTT!!! I DID IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!}}$$
@robjohn I'm the first in the world that did it!!!!!!!!!!
 
Tobias Kildetoft
Tobias Kildetoft
@Chris'ssis Yes (assuming you mean you broke the chat :) )
 
Chris's sis
Chris's sis
It's time to celebrate! Really!!!
@robjohn The only bad thing is that one cannot reduce the whole thing to a very nice form because for a certain trilogarithm there is no known nice closed form. The answer contains a trilogarithm.
@robjohn More precisely $$-4 i \text{Li}_3\left(\frac{1+i}{2}\right)$$
 
robjohn
robjohn
@Chris'ssis is there more, or is that it?
 
Chris's sis
Chris's sis
@robjohn I said this is the part for which the answer cannot be expressed in a nice form.
@robjohn the answer contains 2 more terms (that are neat both).
@robjohn I'M SO INCREDIBLY HAPPY :-)I HAD NOT SO GREAT DAYS, AND NOW LOOK AT IT... :-)
I just go out and howl like a wolf :D
 
Chris's sis
Chris's sis
10:31 AM
@robjohn I didn't note Claude's answer (the closed form) here math.stackexchange.com/questions/937912/…
@robjohn from my work I showed you you can see that my way is completely original. As I said before, all was possible due to the latest research of mine.
@robjohn moreover, I also add it to my book.
 
Alec Teal
Alec Teal
I really want $\bigoplus_{i\in K}^\text{ext}$ to appear with the "ext" bit to the right of the $\oplus$ but the $i\in K$ underneath, like with $\oplus_{i\in K}^\text{ext}$ - how can I do this?
Okay I know it worked there, usually the $i\in K$ is BELOW the circle with the + in it
 
Chris's sis
Chris's sis
@robjohn moreover I know how to precisely tackle the trilogarithm version without any problem. The flow is basically almost the same.
 
Mew
Mew
Hi
anyone on who knows about statistics/probability?
 
Incurrence
Incurrence
11:01 AM
@TedShifrin Why is the exponential map for matrices not injective?
${\bigoplus \limits_{i\in K}^{}}^\text{ext}$
 
teadawg1337
teadawg1337
11:55 AM
Long time no see, chat! ^.^
 
Incurrence
Incurrence
Hey Ttttt Dawwwwg
 
teadawg1337
teadawg1337
@Incurrence Your name looks familiar. Have we met before?
 
Incurrence
Incurrence
<----- Committing to a challenge
 
teadawg1337
teadawg1337
Aaaaaaaaaahhh, that explains why I haven't seen that name in a while
 
Incurrence
Incurrence
Yes, I changed it when a class mate saw my account name
 
ᴇʏᴇs
ᴇʏᴇs
12:03 PM
lol
 
teadawg1337
teadawg1337
I use this username for pretty much everything, lol
 
ᴇʏᴇs
ᴇʏᴇs
They know your website now?
 
Incurrence
Incurrence
@ᴇʏᴇs If they knew I had one
@ᴇʏᴇs They would, but nothing has been mentioned and I have had no surge in Australian views(which would occur)
 
teadawg1337
teadawg1337
Oh wow, I just remembered the following logarithm identity: $$\log_{b}{a}=\frac{\log_d{a}}{\log_d{b}}$$
This would've helped me so much a week ago... I'm facepalming so hard right now....
 
Tobias Kildetoft
Tobias Kildetoft
@AlecTeal To get the indexing placed under the sign, either use \limits or put it inside double dollar signs to get display mode
 
ᴇʏᴇs
ᴇʏᴇs
12:28 PM
@Incurrence Why are your friends in Australia
 
Incurrence
Incurrence
@ᴇʏᴇs What?
 
ᴇʏᴇs
ᴇʏᴇs
@Incurrence I thought you were Canadienne
 
Chris's sis
Chris's sis
@teadawg1337 It's nice to prove that by one single shot. One needs to notice only one thing.
 
Incurrence
Incurrence
@ᴇʏᴇs I am from Queensland - Australia
 
ᴇʏᴇs
ᴇʏᴇs
@Incurrence Oh, are you at University of Queensland
 
Incurrence
Incurrence
12:30 PM
@ᴇʏᴇs There are many unis in Queensland
 
ᴇʏᴇs
ᴇʏᴇs
@Incurrence Is this a comphrensive course offering list or is it incomplete uq.edu.au/study/plan_display.html?acad_plan=MATHEX2320
 
Incurrence
Incurrence
That's the arts department
So very incomplete
That's undergraduate(so 3yrs)
 
ᴇʏᴇs
ᴇʏᴇs
And you have to write a thesis?
 
Incurrence
Incurrence
No
Only if you do 4th year(honors)
 
Tobias Kildetoft
Tobias Kildetoft
@Incurrence You are at Queensland?
 
ᴇʏᴇs
ᴇʏᴇs
12:38 PM
Our school only offers 2-3 of all those courses per semester :(
 
Incurrence
Incurrence
@TobiasKildetoft Yep
 
ᴇʏᴇs
ᴇʏᴇs
Usually closer to 2
 
Tobias Kildetoft
Tobias Kildetoft
I was at a seminar given by a lecturer from there last week (Peter Mcnamara)
 
ᴇʏᴇs
ᴇʏᴇs
We don't have a B.S. in math program
 
iwriteonbananas
iwriteonbananas
12:58 PM
yoyo
@Incurrence no topology? da fuq? T_T
 
Chris's sis
Chris's sis
This day was such a good day to me so far! Hope it keeps like that! This is a really great achievement ... (smashing the whole family of such integrals to a certain extent). An awesome day! :-)
 
Incurrence
Incurrence
@iwriteonbananas Can you help me with a group theory problem?
 
iwriteonbananas
iwriteonbananas
i dont know, what's the problem?
 
Chris's sis
Chris's sis
BBL (I need to buy some food for my lovely pets)
 
Incurrence
Incurrence
I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$
I want to show this is nilpotent
 
Tobias Kildetoft
Tobias Kildetoft
1:09 PM
@Incurrence You mean that $G$ is?
i.e. that that sequence of subgroups terminates at the trivial subgroup
 
Incurrence
Incurrence
$G^0=G$
So I can see that some $g\in G$ has inverse $g^{-1}=2I-g$
 
iwriteonbananas
iwriteonbananas
are you trying to prove that after a finite iteration you arrive at the trivial group?
 
Incurrence
Incurrence
Yep
 
iwriteonbananas
iwriteonbananas
there is a name for that...i cant think of it and it's pissing me off
 
Incurrence
Incurrence
I tried to compute $[G^0,G^0]$ but it went badly
Group nilpotency
$G^n=\{1\}$ for some $n$
Now $[G^0,G^0]=4AB-2ABA-2ABB+ABAB$
From what I can tell, where $A,B\in G$
 
Tobias Kildetoft
Tobias Kildetoft
1:12 PM
@Incurrence It might help to notice that those can be written as $I + A$ with strictly upper triangular $A$, which means that $A^n = 0$ for some $n$
 
Incurrence
Incurrence
Jordan Chevelley?
 
Tobias Kildetoft
Tobias Kildetoft
which helps with expressing the inverses, and might help getting an idea about what the commutators look like (try computing a few specific examples also)
@Incurrence What do you mean?
 
Incurrence
Incurrence
Writing it in such a way is the Jordan-chevalley decomposition
 
Tobias Kildetoft
Tobias Kildetoft
it is? It is just a triviality that they can be written as such
 
Incurrence
Incurrence
Fair enough
 
Tobias Kildetoft
Tobias Kildetoft
1:19 PM
Do you know how to find the inverse of a matrix of the form $I + A$ with nilpotent $A$?
 
Incurrence
Incurrence
Well my inverse here is 2I-A
 
Tobias Kildetoft
Tobias Kildetoft
Ohh, are these $2\times 2$ matrices?
 
Ted Shifrin
Ted Shifrin
@teadawg !!!
hi @Incurrence: Did someone answer your exponential query yet?
 
teadawg1337
teadawg1337
@TedShifrin :D
 
Incurrence
Incurrence
No? Have I screwed up?
@TedShifrin Hi, Nope
 
Tobias Kildetoft
Tobias Kildetoft
1:20 PM
@Incurrence If they are arbitrary, then that will not be the inverse in general
 
Incurrence
Incurrence
@TobiasKildetoft I mean with arbitrary $A\in G$ we have $A^{-1} =2I-A$ right?
 
Ted Shifrin
Ted Shifrin
@Incurrence: The abstract answer is that there are lots of compact matrix groups, so the exponential map on their Lie algebra has to be a covering map. The simplest example is exponentiating real multiples of $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Have you thought about that?
 
Tobias Kildetoft
Tobias Kildetoft
@Incurrence No (try a few examples of $3\times 3$ ones)
 
Ted Shifrin
Ted Shifrin
Glad to see you, @teadawg. You in the middle of finals now?
 
teadawg1337
teadawg1337
@Ted Not yet, I believe they start in the first week of May for me
 
Incurrence
Incurrence
1:23 PM
@TobiasKildetoft Hmm it got closer to $I$, but it didn't inverse
 
Ted Shifrin
Ted Shifrin
Ah, my first one is May 1. ... I was worried about you, since you disappeared, @teadawg. Glad you're doing ok :)
 
Incurrence
Incurrence
@TedShifrin I will think about this, but I don't immediately see it
 
Tobias Kildetoft
Tobias Kildetoft
@Incurrence Do you know the identity for $\frac{1 - x^n}{1-x}$?
 
Ted Shifrin
Ted Shifrin
I'll let you think and we can discuss it later, @Incurrence. Besides, you're busy. :P
 
Incurrence
Incurrence
@TedShifrin Ok
@TobiasKildetoft I can think of expanded forms, but no
 
Tobias Kildetoft
Tobias Kildetoft
1:27 PM
@Incurrence It equals $1 + x + \dots + x^{n-1}$
 
teadawg1337
teadawg1337
@Ted I was slacking a bit in my Psych class, so I had to direct most of my time and energy towards getting caught up. I'm caught up now, so I'm a bit less stressed out :)
 
Ted Shifrin
Ted Shifrin
Ironic that you should stress in a psychology class. Glad you're caught up :)
 
teadawg1337
teadawg1337
Although, it did require me to put off sending the email to the department head at MTSU... I should do that today
 
Ted Shifrin
Ted Shifrin
Ah, I was about to inquire.
 
teadawg1337
teadawg1337
I did a little research, and there doesn't seem to be any professors at MTSU specializing in the analysis of special functions. I have no idea where this will take me
 
Freddy
Freddy
1:33 PM
hello everyone
 
Ted Shifrin
Ted Shifrin
@teadawg: It depends on what people do, of course, but some people working in harmonic analysis or analysis on Lie groups might be interested.
hi @Freddy
 
teadawg1337
teadawg1337
@TedShifrin I just don't know how to think of all the options that are opening up for me... I have no idea where I'll be two years from now, it's all happening so fast...
 
Ted Shifrin
Ted Shifrin
I still say you shouldn't try to get so far ahead of yourself. You should learn stuff methodically and thoroughly, and try to be well-rounded :)
 
teadawg1337
teadawg1337
I plan on diving into differential equations soon, I'm getting a bit tired of working only on integrals and series
 
Ted Shifrin
Ted Shifrin
LOL, ok. There's plenty left to learn ;)
 
Incurrence
Incurrence
2:01 PM
@TedShifrin Is it early btw?
 
Chris's sis
Chris's sis
In my high school textbook there are lots of differential equations.
 
Incurrence
Incurrence
Hmmm you are east US, that's weird
 
Ted Shifrin
Ted Shifrin
LOL, no, @Incurrence, it's 10 AM here. :P
 
Incurrence
Incurrence
@TedShifrin So you normally come here at midday?
 
Ted Shifrin
Ted Shifrin
Why is it weird?
 
Incurrence
Incurrence
2:02 PM
You always turned up at 1:30am[for me], and I assumed it was morning for you lol
 
ᴇʏᴇs
ᴇʏᴇs
West US soon
 
Ted Shifrin
Ted Shifrin
There is no "normally." This is my office hour, but we're on the penultimate day of classes, so I'm not overrun. (Surprisingly, since homework is due in one class in an hour ...)
mr eyeglasses is keeping a close eye ...
 
ᴇʏᴇs
ᴇʏᴇs
I keep a daily log tracking your activities @TedShifrin
 
Ted Shifrin
Ted Shifrin
No, I think your 1:30 AM is closer to our late afternoon.
 
Incurrence
Incurrence
Well it is 1:30 in an hour and a half
 
Ted Shifrin
Ted Shifrin
2:03 PM
You are terribly bored, mr eyeglasses. I guess that's because you're being mistaught complex analysis :(
 
Incurrence
Incurrence
So its 11:30 for you
 
Chris's sis
Chris's sis
@teadawg1337 Not sure your approach is a healthy one. I mean one can learn every day new things, but there is no excuse for saying you get tired of working on integrals and series. There is much stuff there, this is the way it is. A human life is not enough to do research in this area.
 
Incurrence
Incurrence
How is he being mistaught complex?
 
Ted Shifrin
Ted Shifrin
Oh, 11:30 I'm usually in class, but often check in at lunchtime after class, @Incurrence.
 
Incurrence
Incurrence
@TedShifrin So I normally see you past 2am, I am confused lol
Mostly tired though
 
Ted Shifrin
Ted Shifrin
2:04 PM
Not worth being confused, @Incurrence.
 
ᴇʏᴇs
ᴇʏᴇs
I feel worse for the PhD candidates next year; I hope they don't fail their complex qual because of him
 
Incurrence
Incurrence
What is wrong with your complex class?
 
Ted Shifrin
Ted Shifrin
mr eyeglasses, we only know of one problem ...
 
teadawg1337
teadawg1337
@Chris'ssis I'm not going cold turkey on integrals/series, I'm merely adding more variety to my research...
 
ᴇʏᴇs
ᴇʏᴇs
I hope it was just this once
 
Incurrence
Incurrence
2:06 PM
What was the problem?
 
Ted Shifrin
Ted Shifrin
@Incurrence: His professor apparently thinks that all the residues of a meromorphic function at finite poles must add up to $0$. It's only true if one includes the point at infinity.
 
Chris's sis
Chris's sis
@teadawg1337 Sure, this is what I was saying. But never give up working on integrals and series if you wanna reach a top position in this area.
 
Ted Shifrin
Ted Shifrin
Let him learn the rest of mathematics before you commit him to one small subject area, @Chris'ssis.
6
 
teadawg1337
teadawg1337
I don't wish to be the best, I simply wish to enjoy what I do
 
Incurrence
Incurrence
Well normally they work exclusively in the extended complex plane I imagine, did he explicitly say this @ᴇʏᴇs?
 
Chris's sis
Chris's sis
2:07 PM
@TedShifrin Sure, I also encourge him to do that, but not to give up the area I was mentioning (because of some passing feelingS).
 
teadawg1337
teadawg1337
Personally, I'd prefer to excel in many areas over being the best in just one
 
Incurrence
Incurrence
Agreed
 
Chris's sis
Chris's sis
This sounds to me a bit like: "I wanna be the best football player and the best tennis player". I mean the efforts for each one is so amazingly huge.
 
Mats Granvik
Mats Granvik
When I was young I said to my boss at work that I want to be a generalist.
 
teadawg1337
teadawg1337
I don't wish to be the best, I just want to enjoy what I do. I feel the urge to experiment and see other areas I can excel in as well as real/complex analysis
 
Chris's sis
Chris's sis
2:17 PM
@teadawg1337 The aim is not to compare yourself with the others, the aim is to be very good if you wanna enjoy what you do. I mean the hard stuff is often so cool, so you need to be good to attend it.
(well, by practice you get on top)
If you ask me, there are 2 possibilities in the area of integrals, series and limit: to be or not to be like Ramanujan.
It depends on where you wanna remain, and then act accordingly (no matter what the others think about you).
 
teadawg1337
teadawg1337
I don't understand how wanting to be like Ramanujan isn't comparing yourself with others
 
Chris's sis
Chris's sis
@teadawg1337 Well, Ramanujan is a model, but I'm referring to a battle amongst living persons for the best one position. Of course, I cannot think in these terms, no need for such comparisons.
I prefer to learn from others rather than thinking of comparions (when there is something to learn).
I often see solutions far better than mine to some problems, and I appreciate that. Well, I look at that? With much admiration and being willing to learn.
 
teadawg1337
teadawg1337
Sorry if this seems blunt, but I don't see the point you're trying to make here...
 
Chris's sis
Chris's sis
@teadawg1337 The point is that one can desire to be on top without having that craving desire for showing to the world who the best is.
 
evinda
evinda
Hello @DanielFischer !!!
Could I ask you something about Dijkstra?
We have a directed graph G=(V,E), at which each edge $(u,v) \in E$ has a relative value $r(u, v) \in R$ and $ 0 \leq r(u,v) \leq 1$ that represents the reliability at a communication channel, from the vertex u to the vertex v. Consider as r(u, v) the probability that the channel from u to v will not fail the transfer and that the probabilities are independent.

I want to write an efficient algorithm the finds the most reliable path between two vertices, that are given.
 
Chris's sis
Chris's sis
2:33 PM
I bet Ramanujan never cared about who the best in the world was at calculating integrals and series (and many other things from other areas), never interested in comparisons with other people from his time.
 
teadawg1337
teadawg1337
And never am I
 
Daniel Fischer
Daniel Fischer
@evinda Looks OK. You could use vanilla Dijkstra if you let the cost be $-\log r(u,v)$.
 
evinda
evinda
2:47 PM
@DanielFischer Nice :) The line 2 where INITIALIZE-SINGLE-SOURCE(G,s,t) is called requires O(|V|) time, the line 3 requires O(1) time, the line 4 requires O(|V|) time.
Then, the line 6 requires O(|V|) time, the line 7 requires O(1) time. Is it right so far?
How much time does S<-S U {u} require?
Also RELAX requires O(1) time and so the lines 9,10 require $O(\sum_{v \in V} deg(v))=O(E)$ time, right?
 
ADG
ADG
3:01 PM
someone waiting for some fun?
 
Incurrence
Incurrence
@ADG What?
 
ADG
ADG
see this cursors.io
 
Balarka Sen
Balarka Sen
3:30 PM
@MikeMiller I've been meaning to read the proof of Dold-Thom theorem. The $\pi_1 SP^\infty \cong H_1$ case is easy, as $SP^\infty$ is an $H$-space, so $\pi_1$ is abelian. I wonder if something can be done for $\pi_* SP^n$ too.
 
Mike Miller
Mike Miller
3:49 PM
Why should there be, @Balarka?
 
Balarka Sen
Balarka Sen
I don't know. Certainly $\pi_*SP^n$ is not $H_*$ for many spaces. I was just wondering if something can be said about it in general.
 
Mike Miller
Mike Miller
I didn't notice your last sentence. I was referring to what came before.
@BalarkaSen: As usual I don't know what one can say for $\bullet > 1$. Maybe it stabilizes eventually for most spaces or something.
If $\Sigma$ is a genus $g$ surface, $\pi_1(\text{Sym}^g(\Sigma)) \cong H_1(\Sigma)$. Similarly $\pi_2(\text{Sym}^g(\Sigma)) = \pi_2(\Sigma)/\pi_1(\Sigma)$, where here I mean that I'm modding out by the action of $\pi_1$, not a group quotient.
In particular for $g>2, \pi_2(\Sigma) \cong \Bbb Z$.
These are not particularly easy facts.
 
Mats Granvik
Mats Granvik
4:06 PM
Is this right? A root is not the solution to a unique polynomial. A root can be a root to several different polynomials.
 
Mats Granvik
Mats Granvik
4:34 PM
I checked this in Mathematica. A root can be a root to two different polynomials.
 
Remember me
Remember me
Hi @teadawg1337
 
teadawg1337
teadawg1337
Hello @Rememberme
 
Ted Shifrin
Ted Shifrin
4:56 PM
@MatsGranvik I'm not quite sure what the context of your remark is, but of course any number (in whatever field) is a root of uncountably many polynomials.
good night, @Mike
 
Mike Miller
Mike Miller
morning
 
Stan Shunpike
Stan Shunpike
Hey @TedShifrin!
 
Mats Granvik
Mats Granvik
@TedShifrin Ok. Uncountably, there went my hope of counting them.
 
Mike Miller
Mike Miller
@MatsGranvik: You can't count them, but you can say what they all are. Every polynomial over $\Bbb R$ with real root $r$ is of the form $f(x) = (x-x_0)g(x)$, for some other real polynomial $g(x)$.
And vice versa.
 
Chris's sis
Chris's sis
Very nice this question (and especially the answer given by Felix Marin)
3
Q: Integral Involving Dilogarithms

SuperAboundI came across the identity $$\int^x_0\frac{\ln(p+qt)}{r+st}{\rm d}t=\frac{1}{2s}\left[\ln^2{\left(\frac{q}{s}(r+sx)\right)}-\ln^2{\left(\frac{qr}{s}\right)}+2\mathrm{Li}_2\left(\frac{qr-ps}{q(r+sx)}\right)-2\mathrm{Li}_2\left(\frac{qr-ps}{qr}\right)\right]$$ in a book. Unfortunately, as of now, ...

calculus real-analysis integration polylogarithm
 
Remember me
Remember me
5:07 PM
@TedShifrin!!!!
 
Ted Shifrin
Ted Shifrin
hi @Remember
 
Mike Miller
Mike Miller
aaaaaaa i meant real root $x_0$
 
Ted Shifrin
Ted Shifrin
hi @Stan ... Sorry, had students in my office hour. How're you?
 
Stan Shunpike
Stan Shunpike
5:29 PM
I am great! I have been busy with school, but still trying to read study math and physics in my free time. Econ is a lot of fun. I have a great prof this quarter and he's really good about being available to students.
Only thing is.....there are a ridiculous number of typos in the problem sets. I've never seen anything like it.
I've caught at least 6 in 3 sets that were nontrivial.
How are you? I saw you were procrastigrading yesterday lolol
 
Ted Shifrin
Ted Shifrin
I finished the grading, @Stan. One more class left ... ever ... on Monday :)
 
Stan Shunpike
Stan Shunpike
Are you gonna celebrate?
:D
 
Remember me
Remember me
5:47 PM
@TedShifrin the way you have explained Matrix multiplication is amazing......Hats off
salute
 
Stan Shunpike
Stan Shunpike
@Rememberme where did he discuss it? I'd like to read it
 
Remember me
Remember me
Not discuss Videos lectures..... here is the line :-youtube.com/watch?v=b4Ug_1_NtKg
@StanShunpike you like physics it seems so....Have a look at this question physics.stackexchange.com/questions/177780/…
 
Stan Shunpike
Stan Shunpike
@Rememberme cool video. Thx for sharing. Ted is awesome!
 
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