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20:00
@schn when you say that you need $f\in C^{m+1}$, are they specifying an error of $O\!\left(x^{m+1}\right)$ because they want to include the $x^m$ term? That does not require that $f\in C^{m+1}$, but it makes things simpler.
the problem with the show is that everyone has names that sound like they come from middle earth and some of it is in irish.
they should fix that, for me.
20:13
@epsilon-emperor: The problem is to show $\int_T e^{int}\mu(dt)\xrightarrow{n\rightarrow\infty}$. All the acrobatics done in for the hint shows that it is enough to show this for real measures, i.e., enough to assume that $\mu$ is real. Why? because we prove that if $\int_Te^{-int}\mu(dt)$ converges to $0$, then $\int_Te^{-int}f(t)|\mu|(dt)$ converges to $0$ for all bounded (or integrable) $f$.
Have I tampered with some color settings or does everyone see orange galore on Math.SE today ? (new question, OP name, and a few other stuff)...
i'm seeing more orange than usual. but i did just take some mescaline.
kidding.
about the mescaline, not the orange.
You are seeing orange all right. Some dutch soccer fan is corrupting the templates if MSE.
as long as it isn't an irish loyalist thing.
I feel relieved, lol.
20:18
do the loyalists use orange also?
check out meta for even more orange :D
2
it's some reference to a protestant king i think.
Halloween has come early.
my high school wasn't strong on european history, i keep it simple and just learn what colors to hate.
in grad school they had an event for grad students and faculty. i took down all the signs and replaced them with verbatim copies that said NO GANG COLORS in a very large font at the bottom. i would do the same tomorrow.
good plan
20:22
fiddling with signs was a big part of my graduate school education. i can't grow up and i can't change.
i'd replace aspects of the wording with creative variations. i'd change the descriptions of the talks on the seminar schedule. it was a juvenile attempt to shock the bourgeoisie.
if you are educated in or work in a place that has a bulletin board, i recommend this form of behavior as way of jolting people out of the prison of reality.
i don't have this avenue anymore because of work from home.
i remember editing one seminar talk announcement simply by adding the following sentence: "Everyone at this talk is going to be given pills."
i should have been alive in the 60s.
why the 60s?
i think the countercultural spirit that stereotypically animated the times would have been consistent with my own spirit. there was probably bad stuff too.
lyndon b johnson and the beatles i guess
i just wanna fiddle with people's perception of reality.
@leslietownes brilliant
20:32
the cuban missile crisis
mlk's assassination
china blows up its first nuclear bomb
also persistent unremediated institutionalized racism in the united states. yeah ok there was a lot of bad stuff.
elvis returns to the stage
sputnik 1
the problem with these days is nobody wants to goof off and change all of the posters on three floors of a math building. people don't want to put the effort in. they ask, why are you doing that. why would anybody do that.
the french lose in vietnam
i think in the 60s these would not have been questions.
20:34
civil rights
welcome to the age of inertia
where we collectively do nothing as our international passtime
another seminar talk i changed, and i only did this with people giving talks whom i knew, was changed to "We are going to compete to see who can eat the largest amount of crab cakes."
that's what people saw as the description of the talk
the dulles brothers
ah the dear dulles brothers
beginning of the guatemalan civil war
brazilian military coup
another one i did was some variation on "The Buddha teaches joyful participation in the sorrows of life," full stop.
the fact is, because of my methodology i don't know if anyone ever noticed what i was doing.
i certainly never looked at those things except to change them.
20:38
weren't the 70s considered more revolutionary
ah the 70s
chilean military coup
the 70s were more violent and edgy. post altamont the dream died.
vietnam, ofc
beginning of the soviet invasion of afghanistan
argentine military coup
magnavox odyssey
space invaders
iran-iraq war
another summary i changed was from whatever the talk was about, i think sympletic geometry, to "We're alone, we're always alone, and the world is dark and mean."
ms-dos
star wars
pong
atari
20:40
my friend raj gave that talk. i hope someone attended expecting an experience about nihilism.
salvadorean civil war and nicaraguan civil war start
Nixon Shock
shin are you recording a new version of we didn't start the fire
sony walkman
i don't know, i got these recorded in memory
always fun for the perspective
good party trick
Is there some kind of theorem which says that algebraic torus actions on finite dimensional vector spaces have a weight space decomposition?
everything was orange and beige in the 70s and there was a lot of peter max art floating around. that's all i know
sayan i think there is. i'll go downstairs and see where i may have formed this impression.
20:43
@shintuku but, did you live through any of it?
no, i just like history a lot
Yeah if the proof is too algebraic, I don't really care. I just want to build intuition as to why such a statement may be true (it's being used a lot in stuff that I am reading)
i'm not finding it immediately. my classes on representation theory were focused on infinite dimensional noncommutative algebras but i do think there is something.
do you have fell and doran?
What do you mean by a weight space decomposition?
20:48
my cell phone has lost contact with my company's email server and i can't decide if this is something i should talk to IT about or a gift from the universe.
There is a basis $e_i$'s such that the action looks like $t . e_i = t^{r_i} e_i$. Here $r_i$ is a tuple of integers called the weights
on the positive side, darpa gave us the internet :-)
somewhere i have a very funny textbook on networking that has a map of the internet. there are like five points on it.
a physical map, i mean. there's evans hall, something at stanford, something maybe at caltech, and MIT. almost everything.
Atleast for C* that is how the definition goes, I believe it would just be generalized in the same manner for torus actions
@leslietownes Currently I do not, is there a specific place I should look at in it?
20:51
i'll see if i have an infringing copy in my electronic archives. this is ringing some kind of bell but i do not have physical copies of the textbook anymore.
Why isn't this just decomposition of a representation on an abelian group into one dimensional irreps
A torus is abelian
The irreducible characters of $(k^*)^n$ are exactly $(t_1, \cdots, t_n) \mapsto t_i^{n_i}$.
Ah for some reason I thought this was only true of finite abelian groups
5
Q: proof of basic fact that torus actions are diagonalizable

ykmSuppose a torus $T=(\mathbb{C}^\ast)^n$ acts on a finite dimensional vector space $W$, and define for $m \in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by $$W_m = \{w \in W \mid t\cdot w = \chi^m(t)w \text{ for all }t \in T \}$$ i.e. for $w \in W_m$ is a simultaneous eigenvect...

the early academic email setup involved uucp & modems, routing was a nice idea.
Then why is the answer going through all that trouble?
What trouble?
20:59
@OliverDiaz Yep, alright. I'll think more about this.
It seems to just be using simultaneous diagonalization
@Koro Please see this answer. It shows that the proof for $\mathbb{Z}$ also works for $\mathbb{N}$.
Yeah, you just have to observe it's Zariski closure of a certain finitely generated subgroup.
The finitely generated part is simultaneously diagonalizable hence so is the whole thing (the max. torus is Zariski closed)
 
2 hours later…
Bob
Bob
23:03
good evening
Hey, there. Did you still have a question you wanted to ask?
Bob
Bob
well, I cannot get MathJax to render here.
However, it is no big deal.
maybe switch browsers? i have the best luck in chrome.
Bob
Bob
I am currently using Chrome
Thanks for the suggestion, however, I am not going to worry about it
if somebody wanted to look at my post, here is the link:
2
Q: Finding a best fit second order polynomial

BobProblem: Assume we have the following points: $(x_0,y_0), (x_1,y_1), (x_2,y_2), (x_3,y_3)$ where $x_0 = -3$, $x_1 = -2$, $x_2 = -1$ and $x_3 = 0$. Given the function $f(x) = Ax^2 + Bx + C$ find the constants $A$,$B$ and $C$ such that $f(0) = y_3$ and $$d = \sum_{i = 0}^{2} (f(x_i) - y_{i})^2$$ is...

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